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# MEASURE AND INTEGRATION: LECTURE 1 Preliminaries. We

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```MEASURE AND INTEGRATION: LECTURE 1
Preliminaries. We need to know how to measure the “size” or “vol­
ume” of subsets of a space X before we can integrate functions f : X →
R or f : X → C.
We’re familiar with volume in Rn . What about more general spaces
X? We need a measure function µ : {subsets of X} → [0, ∞].
For technical reasons, a measure will not be deﬁned on all subsets of
X, but instead a certain collection of subsets of X called a σ­algebra, a
collection of subsets of X (i.e., a collection M ⊂ P(X) that is a subset
of the power set of X) satisfying the following:
(σ1) X ∈ M.
(σ2) If A ∈ M, then Ac ≡ X \ A ∈ M.
(σ3) If Ai ∈ M (i = 1, 2, . . .), then ∪∞
i=1 ∈ M.
Constrast with a topology τ ⊂ P(X), which satisﬁes
(τ 1) ∅ ∈ τ and X ∈ τ .
(τ 2) If Ui ∈ τ (i = 1, . . . , n), then ∩ni=1 Ui ∈ τ .
(τ 3) If Uα (α ∈ I) is an arbitrary collection in τ , then ∪α∈I Uα ∈ τ .
Remarks on σ­algebras:
(a) By (σ1), X ∈ M, so by (σ2), ∅ ∈ M.
∞
c c
(b) ∩∞
i=1 Ai = (∪i=1 Ai ) ⇒ countable intersections are in M.
(c) A, B ∈ M ⇒ A \ B ∈ M (since A \ B = A ∩ B c ).
Let (X, τX ) and (Y, τY ) be a topological spaces. Then f : X → Y is
continuous if f −1 (U ) ∈ τX for all U ∈ τY . “Inverse images of open sets
are open.”
Let (X, M) be a measure space (i.e., M is a σ­algebra for the space
X). Then f : X → Y is measurable if f −1 (U ) ∈ M for all U ∈ τY .
“Inverse images of open sets are measurable.”
Basic properties of measurable functions.
Proposition 0.1. Let X, Y, Z be topological spaces such that
f
g
X−
→Y −
→ Z.
(1) If f and g are continuous, then g ◦ f is continuous.
Proof. (g ◦ f )−1 (U ) = f −1 (g −1 (U )) = f −1 (open) = open.
Date: September 4, 2003.
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2
MEASURE AND INTEGRATION: LECTURE 1
(2) If f is measurable and g is continuous, then g ◦ f is measurable.
Proof. (g ◦ f )−1 (U ) = f −1 (g −1 (U )) = f −1 (open) = open.
�
Theorem 0.2. Let u : X → R, v : X → R, and Φ : R × R → Y . Set
h(x) = Φ(u(x), v(x)) : X → Y . If u and v are measurable and Φ is
continuous, then h : X → Y is measurable.
Proof. Deﬁne f : X → R × R = R2 by f (x) = u(x) × v(x). Then
h = Φf˙. We just need to show (NTS) that f is measurable. Let
R ⊂ R2 be a rectangle of the form I1 × I2 where each Ii ⊂ R(i = 1, 2) is
an open interval. Then f −1 (R) = u−1 (I1 ) ∩ v −1 (I2 ). Let x ∈ f −1 (R) so
that f (x) ∈ R. Then u(x) ∈ I1 and v(x) ∈ I2 . Since u is measurable,
u−1 (I1 ) ∈ M, and since v is measurable, v −1 (I2 ) ∈ M. Since M is a
σ­algebra, u−1 (I1 ) ∩ v −1 (I2 ) ∈ M. Thus f −1 (R) ∈ M for any rectangle
R. Finally, any open set U = ∪∞
i=1 Ri (rectangle around points with
−1
∞
−1
rational coordinates). So f (U ) = f −1 (∪∞
(Ri ). Each
i=1 Ri ) = ∪i=1 f
term in the union is in M, so since countable unions of elements in M
are in M, f −1 (U ) ∈ M.
�
Examples.
(a) Let f : X → C with f = u + iv and u, v real measurable func­
tions. Then f is complex measurable.
(b) If f = u + iv is complex measurable on X, then u, v, and |f |
are real measurable. Take Φ to be z �→ Re z, z �→ Im z, and
z �→ |z |, respectively.
(c) If f, g are real measurable, then so are f + g and f g. (Also
holds for complex measurable functions.)
(d) If E ⊂ X is measurable (i.e., E ∈ M), then the characteristic
function of E,
�
1 if x ∈ E;
χE (x) =
0 otherwise.
Proposition 0.3. Let F be any collection of subsets of X. Then there
exists a smallest σ­algebra M∗ such that F ⊂ M∗ . We call M ∗ the
σ­algebra generated by F.
Proof. Let Ω = the set of all σ­algebras containing F. The power set
P(X) = the set of all subsets of X is a σ­algebra, so Ω is not empty.
Deﬁne M∗ = ∩M∈Ω M. Since F ∈ M for all M ∈ Ω, we have F ⊂ M∗ .
If M is a σ­algebra containing F, then M∗ ⊂ M by deﬁnition. Claim:
M∗ is a σ­algebra. If A ∈ M∗ , take M ∈ Ω. M is a σ­algebra and
A ∈ M. Thus, Ac ∈ M, and so Ac ∈ M∗ since M∗ ⊂ M. If Ai ∈ M∗
MEASURE AND INTEGRATION: LECTURE 1
3
for each i = 1, 2, . . ., then A ∈ M, and so ∪i Ai ∈ M. It follows that
∪i Ai ∈ M
∗ .
�
Borel Sets. By the previous proposition, if X is a topological space,
then there exists a smallest σ­algebra B containing the open sets. Ele­
ments of B are called Borel sets.
If f : (X, B) → (Y, τ ) and f −1 (U ) ∈ B for all U ∈ τ , then f is
called Borel measurable. In particular, continuous functions are Borel
measurable.
Terminology:
• Fσ (“F­sigma”) = countable union of closed sets.
• Gδ (“G­delta”) = countable intersection of open sets.
MEASURE AND INTEGRATION: LECTURE 2
Proposition 0.1. Let M be a σ­algebra on X, let Y be a topological
space, and let f : X → Y .
(a) Let Ω be a collection of sets E ⊂ Y such that f −1 (E) ∈ M.
Then Ω is a σ­algebra on Y .
(b) If f is measurable and E ⊂ Y is Borel, then f −1 (E) ∈ M.
(c) If Y = [−∞, ∞] (with open sets along with [−∞, a) and (b, ∞]
with a, b ∈ R) and f −1 ((α, ∞]) ∈ M for all α ∈ [−∞, ∞], then
f is measurable.
Proof.
(a) Since f −1 (Y ) = X ∈ M, we have Y ∈ Ω. Also f −1 (E c ) =
(f −1 (E))c ∈ M ⇒ E c ∈ M.
Lastly,
∞
−1
f −1 (∪∞
(Ei ) ∈ M.
i=1 Ei ) = ∪i=1 f
(b) Because f is measurable, all open sets are in Ω. Since Ω is a
σ­algebra, we have B ⊂ Ω.
(c) Recall Ω = {E | f −1 (E) ∈ M}. Given α ∈ R, choose αn < α
so that αn → α as n → ∞. By assumption (αn , ∞] ∈ Ω. Then
∞
c
[−∞, α) = ∪∞
n=1 [−∞, αn ] = ∪n=1 (αn , ∞] . Thus [−∞, α) ∈ Ω.
Then (α, β) = [−∞, β) ∩ (α, ∞] ∈ Ω. Hence, since Ω is a
σ­algebra, Ω contains all open sets. It follows that f is measur­
able.
�
Remark. All of these are equivalent:
f −1
([−∞, α)) ∈ M
⇐⇒ f −1 ([−∞, α]) ∈ M
⇐⇒ f −1 ([α, ∞]) ∈ M
⇐⇒ f −1 ((−∞, α)) & f −1 ({−∞}) ∈ M.
Limits. Let {an } be a sequence in R or [−∞, ∞]. Set
bk = sup{ak , ak+1 , . . .}, k = 1, 2, . . . .
Then inf bk = lim supn→∞ an . As k gets larger, the sup is being
taken over a smaller set, so bk ≥ bk+1 ≥ . . .. Thus bk is a (weakly)
decreasing sequence and so lim bk = inf bk exists. In other words,
Date: September 9, 2003.
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MEASURE AND INTEGRATION: LECTURE 2
lim sup is the largest limit point of the sequence (there exists a sub­
sequence which converges to lim sup). Similarly, we could instead
set bk = inf{ak , ak+1 , . . .} and then sup bk = lim inf n→∞ an , that is,
lim inf is the smallest limit point of the sequence. Note the relation
lim inf an = − lim sup{−an }.
Proposition 0.2. A sequence {an } converges if and only if
lim inf an = lim sup an = lim an .
Limits of functions. Let {fn } : X → R be a sequence of functions.
(sup fn )(x) = sup{fn (x)}
(lim sup fn )(x) = lim fn (x).
n→∞
If, for each x ∈ X, the sequence {fn (x)} converges, then f (x) =
lim fn (x) is the pointwise limit. This works for X = [−∞, ∞] (conver­
gence to ±∞ is obvious).
Theorem 0.3. If, for each i = 1, 2, . . ., the function fi : X →[−∞, ∞]
is measurable, then
g = sup fi and h = lim sup fn
n→∞
i>1
are both measurable.
Proof. NTS g −1 ((α, ∞]) ∈ M for all α. We have
g −1 ((α, ∞]) =
∞
fn−1 ((α, ∞]).
n=1
If x is a member of the LHS, then g(x) > α. Thus, some fn > α from
the deﬁnition of sup. If x is a member of the RHS, then fi (x) > α for
some i, so g ≥ fi (x) > α. Thus, x ∈ g −1 ((α, ∞]). Since fn−1 ((α, ∞]) ∈
M, and since countable unions are in M, g ∈ M. But then lim sup is
measurable as well since by deﬁnition
�
�
lim sup fk = inf sup fk .
j≥1
k≥j
�
Corollary 0.4.
(a) Pointwise limits of measurable functions are
measurable.
(b) If f and g are measurable, then max{f, g} and min{f mg} are
measurable.
MEASURE AND INTEGRATION: LECTURE 2
3
Deﬁne “f plus” and “f minus” as follows:
f + = max{f, 0}, f − = − min{f, 0}.
Then f = f + − f − and |f | = f + + f − .
Simple functions. A simple function is a function that takes only
ﬁnitely many values in R and does not take values ±∞. Let α1 , . . . , αn
be the values and Ai = {x ∈ X | s(x) = αi }. Then
s=
n
�
αi χAi .
i=1
Note: χAi is measurable ⇐⇒ Ai ∈ M. Constant function αi is
measurable ⇒ product αi χAi is measurable. Since sums of measurable
functions are measurable, s is measurable ⇐⇒ all Ai are measurable.
Theorem 0.5. Let f : X → [0, ∞] be measurable. Then there exists a
sequence 0 ≤ s1 ≤ s2 ≤ . . . of measurable simple functions such that
limn→∞ sn = f .
Proof. Partition [0, n] into intervals of length 2−n . Deﬁne ϕn : [0, ∞] →
[0, ∞) as follows. Let δn = 2−n . For each t, choose kn (t) such that
kδn ≤ t < (k + 1)δn . Put
�
kn (t)δn
0 ≤ t < n;
ϕn (t) =
n
n ≤ t ≤ ∞.
Note that each ϕn is Borel measurable, ϕ1 ≤ ϕ2 ≤ · · · ≤ t and
limn→∞ ϕn (t) = t. Let sn = ϕn ◦ f . Then for any open set U ,
−1 −1
ϕn (U ) = f −1 (Borel set) ∈ M. Thus sn is measurable,
s−1
n (U ) = f
increasing, and its limit is f .
�
A positive measure is a mapping µ : M → [0, ∞] which is countably
∞
�
∞
µ (∪i=1 Ai ) =
µ(Ai ) for disjoint Ai .
i=1
Lebesgue Integral. Let (X, M, µ) be a measure space. Let s : X →
�
[0, ∞) be the simple function s = N
i=1 αi χAi , where the Ai are disjoint.
For each E ∈ M, deﬁne the integral
�
N
�
sdµ =
αi µ(Ai ∩ E).
E
i=1
4
MEASURE AND INTEGRATION: LECTURE 2
For more general measurable functions f : X → [0, ∞], then
��
�
�
f dµ = sup
sdµ | s simple, 0 ≤ s ≤ f .
E
E
If f : X → [−∞, ∞], then
�
�
�
+
f dµ =
f dµ −
f − dµ,
E
E
E
provided both terms on the right­hand side are ﬁnite.
MEASURE AND INTEGRATION: LECTURE 3
Riemann integral. If s is simple and measurable then
�
N
�
sdµ =
αi µ(Ei ),
X
�N
i=1
where s = i=1 αi χEi . If f ≥ 0, then
��
�
�
f dµ = sup
sdµ | 0 ≤ s ≤ f, s simple & measurable .
X
X
Recall the Riemann integral of function f on interval [a, b]. Deﬁne
lower and upper integrals L(f, P ) and U (f, P ), where P is a partition
of [a, b]. Set
�
�−
f = sup L(f, P ) and
f = inf U (f, P ).
−
P
P
A function f is Riemann integrable ⇐⇒
�
�−
f = f,
−
�
in which case this common value is f .
A set B ⊂ R has measure zero if, for any � > 0, there exists a
∞
countable collection of intervals {Ii }∞
i=1 such that B ⊂ ∪i=1 Ii and
�
∞
i=1 λ(Ii ) < �. Examples: ﬁnite sets, countable sets. There are also
uncountable sets with measure zero. However, any interval does not
have measure zero.
Theorem 0.1. A function f is Riemann integrable if and only if f is
discontinuous on a set of measure zero.
A function is said to have a property (e.g., continuous) almost ev­
erywhere (abbreviated a.e.) if the set on which the property does not
hold has measure zero. Thus, the statement of the theorem is that f is
Riemann integrable if and only if it is continuous almost everywhere.
Date: September 11, 2003.
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MEASURE AND INTEGRATION: LECTURE 3
Recall positive measure:
a measure function µ : M → [0, ∞] such
�∞
that µ (∪∞
E
)
=
µ(E
i ) for Ei ∈ M disjoint.
i=1 i
i=1
Examples.
(1) “Counting measure.” Let X be any set and M = P(X) the set
of all subsets. If E ⊂ X is ﬁnite, then µ(E) = #E (the number
of elements in E). If E ⊂ X is inﬁnite, then µ(E) = ∞.
(2) “Unit mass at x0 – Dirac delta function.” Again let X be any
set and M = P(X). Choose x0 ∈ X. Set
�
1
if x0 ∈ E;
µ(E) =
0
if x0 �∈ E.
Theorem 0.2.
(1) If E ⊂ R and µ(E) < ∞, then µ(∅) = 0.
(2) (Monotonicity) A ⊂ B ⇒ µ(A) ≤ µ(B).
(3) If Ai ∈ M for i = 1, 2, . . ., A1 ⊂ A2 ⊂ · · · , and A = ∪∞
i=1 Ai ,
then µ(Ai ) → µ(A) as i → ∞.
(4) If Ai ∈ M for i = 1, 2, . . ., A1 ⊃ A2 ⊃ · · · , µ(A1 ) < ∞, and
A = ∩∞
i=1 Ai , then µ(Ai ) → µ(A) as i → ∞.
Proof.
(1) E = E ∪ ∅ ⇒ µ(E) = µ(E) + µ(∅).
(2) B = A ∪ (B \ A) ⇒ µ(B) = µ(A) + µ(B \ A) ≥ µ(A).
(3) Let B1 = A1 , B2 = A2 \ A1 , B3 = A3 \ A2 , . . .. Then the
Bi are disjoint, An = B1 ∪ · · · ∪�Bn , and A = ∪∞
i=1 Bi . Thus,
n
µ(An ) = µ(B1 ) + · · · + µ(Bn ) = i=1 µ(Bi ), and (3) follows.
(4) Let Cn = A1 \ An . Then C1 ⊂ C2 ⊂ · · · . We have λ(Cn ) =
λ(A1 ) − λ(An ). Also, A1 \ A = ∪n Cn . Thus, A1 ∩ (∩Ai )c =
∪(A1 \ An ), and so
µ(A1 \ A) = lim µ(Ci ) = µ(A1 ) − lim µ(An ).
i→∞
Hence, µ(A1 ) → µ(A).
i→∞
�
Properties of the Integral.
�
�
(a) If 0 ≤ f ≤ g on E, then E f dµ ≤ E gdµ.
�
�
(b) If A ⊂ B, A, B ∈ M, and f ≥ 0, then A f dµ ≤ B f dµ.
(c) If
� f ≥ 0 and
� c ∈ [0, ∞) is a non­negative constant, then
cf
dµ
=
c
f dµ.
E
E
�
(d) If f (x) = 0 for all �x ∈ E, then E f dµ = 0.
(e) If µ(E) = 0, then
f dµ�= 0.
E
�
(f) If f ≥ 0, then E f dµ = E χE f dµ.
Proof.
(a) If s ≤ f is simple, then s ≤ g so the sup on g is over a
larger class of simple functions than the sup on f .
MEASURE AND INTEGRATION: LECTURE 3
3
(b) We have Ei ∩ A ⊂ Ei ∩ B for all Ei . If s is simple,
�
�
N
N
�
�
sdµ =
αi µ(Ei ∩ A) ≤
αi µ(Ei ∩ B) =
sdµ.
A
i=1
(c) For any simple s,
�
B
i=1
�
E
csdµ = c
(cαi )χEi
i
�
sdµ since
E
�
=c
αi χEi .
i
For any constant c, s ≤ f ⇐⇒ cs ≤ cf . Thus,
�
�
�
�
�
�
cf = sup s = sup
s = sup cs = c f.
s≤cf
s/c≤f
s� f
�N
(d) Let s ≤ f be simple and s =
i=1 αi χEi . Without loss of
generality, α1 = 0 and E1 ⊃ E. Thus,
�
N
�
sdµ =
αi µ(Ei ∩ E) = α1 µ(E) = 0.
E
i=1
(The convention here
is�
that 0 · ∞ = 0.)
�
� and throughout
(e) If s ≤ f and s = i αi χEi , then E s = i αi µ(E ∩ Ei ) = 0.
(f) This could have been the deﬁnition of the integral.
�
MEASURE AND INTEGRATION: LECTURE 4
Integral is additive for simple functions.
Proposition� 0.1. Let s and� t be non­negative
measurable simple func­
�
tions. Then X (s + t)dµ = X s dµ + X t dµ.
�
Proof. Let E ∈ M and deﬁne ϕ(E) = E s dµ. First we show that ϕ is
measurable. To this end, let Ei ∈ M with Ei disjoint and E = ∪∞
i=1 Ei .
Then
N
�
ϕ(E) =
αi µ(Ai ∩ E) =
i=1
N
�
=
N
�
�
�
αi µ Ai ∩ (∪∞
j=1 Ej )
i=1
αi µ (∪∞
i=1 (Ai
∩ Ej )) =
i=1
Let s =
�N
i=1
αi
i=1
∞ �
N
�
=
N
�
αi µ(Ai ∩ Ej ) =
j=1 i=1
∞ �
�
j=1
Ej
∞
�
µ (Ai ∩ Ej )
j=1
s dµ =
∞
�
ϕ(Ej )
j=1
�
αi χAi and t = M
j=1 βj χBj . Let Eij = Ai ∩ Bj . Then
�
(s + t)dµ = (αi + βj )µ(Eij )
Eij
and
�
�
Eij
so that
s dµ +
Eij
�
t dµ = αi µ(Eij ) + βj µ(Eij )
�
Eij
s dµ +
�
Eij
t dµ =
Eij
(s + t)dµ.
Write X = ∪i,j Eij as a disjoint union. Then
�
�
�
s dµ +
t dµ =
(s + t)dµ.
X
X
X
�
Date: September 16, 2003.
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MEASURE AND INTEGRATION: LECTURE 4
Next, we want to prove
�
�
�
(f + g)dµ =
f dµ +
g dµ.
X
X
X
If s ≤ f simple and t ≤ g simple, then s + t ≤ f + g simple. The only
thing we know so far is that
�
�
�
(f + g)dµ ≥
f dµ +
g dµ.
X
X
X
One way to obtain equality is to deﬁne an upper integral and a lower
integral, and say that a function is integrable ⇐⇒ its upper and lower
integral are equal and ﬁnite. Then we should prove that f integrable
⇐⇒ f measurable. But this is not necessary, and we will use the
deﬁnition we have.
Monotone convergence.
Theorem 0.2. Let fn : X → [0, ∞] be a sequence of measurable func­
tions such that
(a) 0 ≤ f1 ≤ f2 ≤ · · · ≤ ∞, and
(b) fn (x) → f (x) as n → ∞ for all x ∈ X.
�
�
Then X fn dµ → X f dµ as n → ∞.
�
�
�
Proof. Since fi ≤ fi+1 for all i, we have
f
≤
f
.
Thus
fi → α
i
i+1
�
�
�
for some α ∈ [0, ∞]. Also fn ≤ f ⇒ fi ≤ f , so α ≤ f . Next, let
s be simple and measurable with 0 ≤ s ≤ f and let c be a constant
such that 0 ≤ c ≤ 1. Deﬁne En = {x | fn (x) ≥ cs(x)} for n = 1, 2, . . ..
Then Ei is measurable and E1 ⊂ E2 ⊂ · · · , and X = ∪∞
i=1 Ei . Indeed,
if f (x) = 0 for any x ∈ X, then x ∈ E1 , and if f (x) > 0, then
cs(x) < f (x). Since fn → f , fn > cs(x) for n large; thus x ∈ En for n
large.
Lastly,
�
�
�
Letting n → ∞,
X
fn dµ ≥
En
fn dµ ≥ c
En
s dµ.
�
α ≥ c lim
n→∞
En
s dµ.
�
�
Thus α ≥ c X sdµ for any c < 1. Let c → 1. Then α ≥ X s dµ for
any simple, measurable 0 ≤ s ≤ f . We conclude that
�
α=
f dµ = lim fn dµ.
X
n→∞
�
MEASURE AND INTEGRATION: LECTURE 4
3
Integral is additive for non­negative measurable functions.
Theorem 0.3. Let
func­
� a sequence
�
� fn : X → [0, ∞] be
�∞of measurable
tions and f (x) = ∞
f
(x).
Then
f
dµ
=
f
dµ.
n=1 n
n=1 X n
X
�
�
�
Proof. First, claim if f, g measurable, then X f + g = X f + X g. Let
0 ≤ s1 ≤ s2 ≤ · · · be simple, measurable, and si → f . Similarly, let
0 ≤ t1 ≤ t2 ≤ · · · be simple, measurable, and ti → �g. Then si + ti
are
� simple �and si + ti → f + g, which implies that X (si + ti )dµ =
s dµ + X ti dµ. By the monotone
convergence
theorem, the claim
X i
�
� ��N
�N �
is proved. Using induction, X
i=1 fi dµ =
i=1 X fi dµ.
�N
�∞
Let gN = i=1 fi . Then gN → n=1 fn = f (x) as N → ∞ mono­
tonically. Thus
� �
�
∞
fn dµ = lim
gN dµ
N →∞
X n=1
X
� �
N
= lim
N →∞
X i=1
N �
�
= lim
N →∞
=
∞ �
�
i=1
i=1
X
X
fi dµ
fi dµ
fi dµ
�
Interchanging summation and integration.
Corollary 0.4. Let X = Z+ ≡ {1, 2, 3, . . .} and µ be the counting
measure. Let aij ≥ 0 and fj = aij : Z+ → [0, ∞]. Then
� �
∞
∞ �
�
fj =
fj ,
j=1
so that
∞ �
∞
�
i=1 j=1
j=1
aij =
∞ �
∞
�
aij .
j=1 i=1
Fatou’s Lemma.
Lemma 0.5. Let fn : X → [0, ∞] be a sequence of measurable func­
tions. Then
� �
�
�
lim inf fn dµ ≤ lim inf
fn dµ.
X
n→∞
n→∞
X
4
MEASURE AND INTEGRATION: LECTURE 4
Proof. Let gk (x) = inf i≥k fi (x). Then g1 ≤ g2 ≤ · · · and
lim gk = lim inf fk .
k→∞
k→∞
Also, gk ≤ fk , so monotone convergence implies that
�
�
lim inf fk dµ =
lim gk dµ
X
X
�
= lim
gk dµ
X
�
= lim inf
gk dµ
X
�
≤ lim inf
fk dµ.
X
�
MEASURE AND INTEGRATION: LECTURE 5
Deﬁnition of L1 . Let f : X → [−∞, ∞] be measurable.
� We say that
f is �in L1 (written f ∈�L1 (µ) or simply f ∈ L1 ) ⇐⇒ X f + dµ < ∞
and X f − dµ∞ ⇐⇒ X |f | dµ < ∞. Deﬁne
�
�
�
+
f dµ =
f dµ −
f − dµ
X
X
X
when at least one of the terms on the right­hand side is ﬁnite.
Integral of complex functions. Let f : X → C be measurable. That
is, f = u + iv where u, v : X →
R are measurable. Then
�
1
f ∈ L (µ) ⇐⇒
|F | dµ < ∞
X
�
�
⇐⇒
|u| dµ < ∞ and
|v | dµ < ∞.
X
Deﬁne
�
X
�
f dµ =
X
�
=
X
X
�
u dµ + i
+
X
�
u dµ −
X
v dµ
−
u dµ + i
�
X
�
+
v dµ − i
X
v − dµ.
1
Theorem 0.1. Let f, g ∈ L (µ). If α, β ∈ C, then αf + βg ∈ L1 (µ)
and
�
�
�
(αf + βg)dµ = α
f dµ + β
g dµ.
X
X
X
Proof. First, αf + βg is measurable, and by the triangle inequality,
|αf + βg| ≤ |α| |f | + |β| |g |, so
�
�
�
|αf + βg| ≤ |α|
|f | + |β|
|g | < ∞.
X
X
X
Just need to show that
�
�
�
(1) �X (f + g)dµ =� X f dµ + X g dµ, and
(2) X (αf )dµ = α X f dµ.
Date: September 18, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 5
For (1), assume f, g real; the complex case follows from the real case.
Let h = f + g. Then h+ − h− = f + − f − + g + − g − , so h+ + f − + g − =
f + + g + + h− . Since the integral is linear for non­negative functions,
�
�
�
�
�
�
+
−
−
+
+
h + f + g = f + g + h− ⇒
�
�
�
�
�
�
+
−
+
−
+
h − h = f − f + g − g−.
�
�
�
Thus, f + g = f + g.
For (2), let α = a + bi for a, b ∈ R. Then
�
�
�
αf = (a + bi)(u + iv) = au + aiv + biu − bv
�
= (au − bv + i(av + bu)
�
�
= (au − bv) + i (av + bu).
Also,
��
�
�
�
(a + bi)
(u + iv) = (a + bi)
u+i v
�
�
�
�
= a u + bi u + ai v − b v.
�
�
So, just need to show that au = a u. If a = 0, then both sides
vanish. If a > 0, then
�
�
�
+
(au) = (au) − (au)−
�
�
+
= a · u − a · u−
�
�
�
+
−
= a u − a u = a u.
If a < 0, then
�
�
�
(au)−
�
�
−
= (−a) · u − (−a) · u+
�
�
−
= −a u − (−a) u+
��
�
�
�
+
−
=a
u − u
= a u.
au =
+
(au) −
MEASURE AND INTEGRATION: LECTURE 5
3
�
Theorem 0.2. If f ∈ L1 (µ), then
��
� �
�
�
� f dµ� ≤
|f | dµ.
�
�
X
X
Proof. For some θ ∈ [0, 2π),
��
�
�
�
�
�
f dµ = �
f dµ��
eiθ .
X
Hence,
X
��
�
�
�
�
�
� f dµ� = e−iθ
f dµ =
(e−iθ f )dµ
�
�
X
X
X
��
�
−iθ
= Re
e f dµ
X
�
=
Re(e−iθ f )dµ
�X
�
� −iθ �
�
�
≤
e f dµ =
|f | dµ.
X
X
�
Dominated convergence.
Theorem 0.3. Let fn : X → C be a sequence of measurable functions,
and assume that f (x) = limn→∞ fn (x) (that is, the sequence fn con­
verges pointwise). If there exists g ∈ L1 (µ) such that |fn (x)| ≤ g(x)
for all n and for all x ∈ X, then f ∈ L1 (µ) and
�
�
�
lim
|fn − f | dµ = 0, so lim
fn dµ =
f dµ.
n→∞
n→∞
X
X
X
Proof. Since |fn (x)| ≤ g(x) for all�n, the limit
� f has the property 1that
|f | ≤ |g(x)|. This means that |f | ≤ |g | < ∞,
so f ∈ L (µ).
Next, |fn − f | ≤ |fn | + |f | ≤ 2g, which means that 2g − |fn − f | ≥ 0.
Applying Fatou’s lemma,
�
�
2g dµ ≤ lim inf (2g −
|fn − f |)dµ
X
X
�
�
=
2g dµ + lim inf
− |fn − f | dµ
X
�
�X
=
2g dµ − lim sup
|fn − f | dµ.
X
X
4
Since
�
X
MEASURE AND INTEGRATION: LECTURE 5
2g < ∞, it can be cancelled from both sides. Thus,
�
lim sup
|fn − f | dµ ≤ 0,
X
and so
�
lim
X
|fn − f | dµ = 0.
From the previous theorem,
��
� �
�
�
� (fn − f )dµ� ≤
|fn − f | dµ
�
�
X
X
��
� �
�
�
�
⇒ ��
fn dµ −
f dµ�� ≤
|fn − f | dµ → 0
X
X
X
�
�
⇒ fn dµ →
f dµ.
X
X
�
Sets of measure zero. Let (X, M, µ) be a measure space and E ∈
M. A set E has measure zero if and only if µ(E) = 0. If f, g : X → C,
then f = g almost everywhere (a.e.) if N = {x | f (x) = g(x)} has
measure zero. Deﬁne an equivalence relation f ∼ g if f = g a.e.
�
�
Proposition 0.4. If f ∼ g, then, for all E ∈ M, E f dµ = E g dµ.
Proof. Write E as disjoint union E = (E \ N ) ∪ (E ∩ N ). Then, since
f = g away from N , and since N has measure zero,
�
�
�
f dµ =
f dµ +
f dµ
E
E\N
E∩N
�
�
=
g dµ + 0 =
g dµ.
E\N
E
�
Completion of a σ­algebra.
Theorem 0.5. Let (X, M, µ) be a measure space. Let
M∗ = {E ⊂ X | ∃A, B ∈ M : A ⊂ E ⊂ B & µ(B \ A) = 0}.
Now deﬁne µ(E) = µ(A) for all E ∈ M∗ . Then M∗ is a σ­algebra and
this deﬁnition of µ is a measure.
The measure space (X, M∗ , µ) is a called the completion of the mea­
sure space (X, M, µ). A measure space is complete if it is equal to its
completion.
Note.
If f is only deﬁned a.e. (say, except for a�set N of measure
zero), then we can deﬁne f (x) = 0 for all x ∈ N . ⇒ f is well deﬁned.
MEASURE AND INTEGRATION: LECTURE 5
5
Theorem
(a) Let f : X → [0, ∞] be measurable, E ∈ M, and
� 0.6.
f
dµ
=
0.
Then �f = 0 a.e. on E.
E
1
(b) Let f ∈ L (µ) and E f dµ = 0 for every E ∈ M. Then f = 0
a.e. on X.
Proof.
(a) Let An = {x ∈ E | f (x) > 1/n}. Then
�
�
�
1
f dµ ≥
f dµ ≥
1/n dµ = µ(An ),
n
E
An
An
which implies that µ(An )�
= 0. But {x | f (x) > 0} = ∪∞
i=1 An
and µ({x | f (x) > 0}) ≤ ∞
µ
(A
)
=
0.
n
i=1
�
(b) Let
f
=
u
+
iv.
Choose
E
=
{
x
|
u
(
x
)
≥
0
}
.
Then
f =
E
�
� +
� +
+
u + i E v ⇒ E u = 0 and by (a), u = 0 a.e.
E
�
�∞
Theorem 0.7. Let Ek ∈ M such that k=1 µ(Ek ) < ∞. Then almost
every x ∈ X lie in at most ﬁnitely many Ek .
Proof. Let
�A = {x ∈ X | x ∈ Ek for inﬁnitely many k}. NTS µ(A) = 0.
Let g = ∞
i=1 χEk .
Then x ∈ A ⇐⇒ g(x) = ∞. We have
�
∞ �
∞
�
�
g dµ =
χEk dµ =
µ(Ek ) < ∞.
X
i=1
1
X
i=1
In other words, g ∈ L (µ) and thus g(x) < ∞ a.e.
�
MEASURE AND INTEGRATION: LECTURE 6
Lebesgue measure on Rn . We will deﬁne the Lebesgue measure
λ : {subsets of Rn } → [0, ∞] through a series of steps.
(1) λ(∅) = 0.
(2) Special rectangles: rectangles with sides parallel to axes.
• n = 1: λ([a, b]) = b − a
• n = 2: λ([a1 , b1 ] × [a2 , b2 ]) = (b1 − a1 )(b2 − a2 ).
..
.
(3) Special polygons: ﬁnite unions of special rectangles. To ﬁnd
measure, write P = ∪N
k=1 Ik , Ik disjoint special rectangles. De­
�N
ﬁne λ(P ) = k=1 λ(Ik ).
Properties of Lebesgue measure.
�N
N�
�
(1) Well­deﬁned. If P = ∪N
k=1 Ik = ∪k=1 Ik , then
k=1 λ(Ik ) =
�N �
�
k=1 λ(Ik ). (Exercise)
(2) P1 ⊂ P2 ⇒ λ(P1 ) ≤ λ(P2 ).
(3) P1 , P2 disjoint ⇒ λ(P1 ∪ P2 ) = λ(P1 ) + λ(P2 ).
Open sets. Let G ⊂ Rn be open and nonempty. We will approximate
G from within by special polygons. That is, we deﬁne
λ(G) = sup{λ(P ) | P ⊂ G, P special polygon}.
Properties for open sets.
(1) λ(G) = 0 ⇐⇒ G = ∅. (Nontrivial open sets have positive
measure.)
(2) λ(Rn ) = ∞.
(3) G1 ⊂ G2 ⇒ λ(G
�1∞) ≤ λ(G2 ).
∞
(4) λ (∪k=1 Gk ) ≤ k=1 λ(Gk ).
�∞
(5) Gk open and pairwise disjoint ⇒ λ (∪∞
k=1 Gk ) =
k=1 λ(Gk ).
◦
◦
(6) P special polygon ⇒ λ(P ) = λ(P ), where P = interior of P .
Proof.
(3) If P ⊂ G1 , then P ⊂ G2 . Thus λ(P ) ≤ λ(G2 ). Taking
sup over all special polygons P gives the desired result.
Date: September 23, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 6
N
(5) Let P ⊂ ∪∞
k=1 Gk . Claim: can write P = ∪k=1 Pk� with Pk
special polygons, Pk ⊂ Gk� and Pk not contained in any other
Gk . Then
λ(P ) =
N
�
λ(Pk� ) ≤
k=1
N
�
λ(Gk� ) ≤
k� =1
∞
�
λ(Gk ).
k=1
�∞
Taking sup over all P , λ (∪∞
k=1 Gk ) ≤
k=1 λ(Gk ).
(6) Fix N and choose P1 , . . . , PN special polygons such that Pk ⊂
N
∞
Gk . Then Pk ’s disjoint ⇒ ∪N
k=1 Pk ⊂ ∪k=1 Gk ⊂ ∪k=1 Gk . Thus,
�N
�
�∞
�
N
�
λ(Pk ) = λ
Pk ≤ λ
Gk .
k=1
k=1
k=1
�N
Taking sup over all P1 , . . . , PN , k=1 λ(Gk ) ≤ λ (∪∞
k=1 Gk ). Let­
ting N → ∞,
�∞
�
∞
�
λ(Gk ) ≤ λ
Gk .
k=1
k=1
The reverse inequality is simply (5), and so equality must hold.
(7) Clearly, for any � > 0 we can ﬁnd P � ⊂ P ◦ such that λ(P � ) >
λ(P ) − �. Thus,
λ(P ) − � < λ(P � ) ≤ λ(P ◦ ),
and so λ(P ) ≤ λ(P ◦ ). Of course, the inequality λ(P ◦ ) ≤ λ(P )
also holds, because, if Q ⊂ P ◦ is a special polygon, then λ(Q) ≤
λ(P ) and we simply take sup over all such Q.
�
Compact sets. Let K ⊂ Rn . We will approximate K by open sets.
That is, we deﬁne
λ(K) = inf{λ(G) | K ⊂ G, G open}.
Claim: the deﬁnition is well­deﬁned. (In particular, a special polygon
P is compact.)
Proof. Let α = old λ(P ) and β = new λ(P ). If P ⊂ G, then λ(P ) ≤
λ(G), so by taking inf over all G, α ≤ β. For the other inequality,
�
� ◦
say P = ∪N
k=1 Ik . Choose Ik larger than Ik so that (Ik ) ⊃ Ik and
� ◦
λ(Ik� ) < λ(Ik ) + �/N for some ﬁxed � > 0. Let G = ∪N
k=1 (Ik ) . Then
MEASURE AND INTEGRATION: LECTURE 6
3
P ⊂ G and G is open. We have
β ≤ λ(G) ≤
N
�
λ ((Ik� )◦ ))
k=1
=
N
�
λ(Ik� )
k=1
<
N
�
λ(Ik ) + �/N
k=1
= α + �.
Since this is true for any � > 0, β ≤ α, and consequently α = β.
�
Properties for compact sets.
(1) 0 ≤ λ(K) < ∞.
(2) K1 ⊂ K2 ⇒ λ(K1 ) ≤ λ(K2 ).
(3) λ(K1 ∪ K2 ) ≤ λ(K1 ) + λ(K2 ).
(4) If K1 and K2 are disjoint, then λ(K1 ∪ K2 ) = λ(K1 ) + λ(K2 ).
Proof.
(2) If K2 ⊂ G (G open) then K1 ⊂ G.
(3) If K1 ⊂ G1 and K2 ⊂ G2 , then K1 ∪ K2 ⊂ G1 ∪ G2 . Thus,
λ(K1 ∪ K2 ) ≤ λ(G1 ∪ G2 ) ≤ λ(G1 ) + λ(G2 ). Take inf over all
G1 , G2 ⇒ λ(K1 ∪ K2 ) ≤ λ(K1 ) + λ(K2 ).
(4) Since K1 and K2 are compact (and disjoint), there exists � > 0
such that an �­neighborhood K1� of K1 does not intersect K2
and an �­neighborhood K2� of K2 does not intersect K1 . Let
G be an open set such that K1 ∪ K2 ⊂ G. Let G1 = G ∩ K1�
and G2 = G ∩ K2� . Then G1 and G2 are disjoint, Ki ⊂ Gi for
i = 1, 2, and
λ(K1 ) + λ(K2 ) ≤ λ(G1 ) + λ(G2 ) = λ(G1 ∪ G2 ) ≤ λ(G).
Taking inf over all G gives λ(K1 ) + λ(K2 ) ≤ λ(K1 ∪ K2 ). The
reverse inequality is (3) ⇒ λ(K1 ∪ K2 ) = λ(K1 ) + λ(K2 ).
�
Inner and outer measure. If A ⊂ Rn is arbitrary, then we deﬁne
both inner and outer measure:
• (Outer measure) λ∗ (A) = inf{λ(G) | A ⊂ G, G open}.
• (Inner measure) λ∗ (A) = sup{λ(K) | K ⊂ A, K compact}.
Properties:
(1) λ∗ (A) ≤ λ∗ (A).
(2) A ⊂ B ⇒ λ∗ (A) ≤ λ∗ (B) and λ∗ (A) ≤ λ∗ (B).
4
MEASURE AND INTEGRATION: LECTURE 6
�∞ ∗
(3) λ∗ (∪∞
k=1 Ak ≤
k=1 λ (Ak ). (Outer measure is countably sub­
�∞
(4) If Ak disjoint, then λ∗ (∪∞
k=1 Ak ) ≥
k=1 λ∗ (Ak ).
(5) If A open or compact, then λ∗ (A) = λ∗ (A) = λ(A).
Proof.
(1) If K ⊂ A ⊂ G, then K ⊂ G, so λ(K) ≤ λ(G) by
deﬁnition of λ(compact).
(3) For any � > 0, choose Gk ⊃ Ak such that λ(Gk ) < λ∗ (Ak )+�/2k .
Then
�∞
�
�∞
�
∞
�
∗
λ
Ak ≤ λ
Gk ≤
λ(Gk )
k=1
k=1
<
k=1
∞
�
�
λ∗ (Ak ) + �/2k
�
k=1
=
∞
�
λ∗ (Ak ) + �.
k=1
(4) Choose Kk ⊂ Ak ; then Kk ’s disjoint. Then
�∞
�
�N
�
N
�
λ∗
Ak ≥ λ
Kk =
λ(Kk ).
k=1
k=1
k=1
With N ﬁxed, take sup over all Kk . This gives
�∞
�
∞
�
λ∗
Ak ≥
λ∗ (Ak ).
k=1
k=1
Letting N → ∞ gives the result.
(5) First let A be open. Then λ∗ (A) = λ(A). If P ⊂ A with P
special polygon, then λ(P ) ≤ λ∗ (A), which implies that λ(A) ≤
λ∗ (A). Thus,
λ∗ (A) = λ(A) ≤ λ∗ (A) ≤ λ∗ (A),
so all are equal. Now let A be compact. Then λ∗ (A) = λ(A),
and λ(A) = λ∗ (A) since the measure of compact sets was de­
ﬁned using open sets.
�
MEASURE AND INTEGRATION: LECTURE 7
Review. The steps to deﬁning Lebesgue measure. (1) measure of
rectangles (2) measure of special polygons (3) measure of open sets:
λ(G) = sup{λ(P ) | P ⊂ G, P special polygon}. (4) measure of com­
pact sets: λ(K) = inf{λ(G) | K ⊂ G, G open}. (5) Inner λ∗ and outer
λ∗ measures.
Lebesgue measurable sets (with ﬁnite outer measure). Let A ⊂
Rn and λ∗ (A) < ∞ (A has ﬁnite outer measure). Then we write that
A ∈ L0 ⇐⇒ λ∗ (A) = λ∗ (A) and deﬁne measure of A to be
λ(A) = λ∗ (A) = λ∗ (A).
We know that L0 contains all compact sets and open sets of ﬁnite
measure.
Lemma 0.1. Let A, B ∈ L0 . If A and B are disjoint, then A ∪ B ∈ L0
and λ(A ∪ B) = λ(A) + λ(B).
Proof.
λ∗ (A ∪ B) ≤ λ∗ (A) + λ∗ (B)
= λ(A) + λ(B)
(A, B ∈ L0 )
= λ∗ (A) + λ∗ (B)
(Property of inner measure)
∗
≤ λ∗ (A ∪ B) ≤ λ (A ∪ B)
�
Main approximation theorem.
Theorem 0.2. Let A ∈ Rn and λ∗ (A) < ∞. Then A ∈ L0 if and
only if for all � > 0 there exists K compact and G open such that
K ⊂ A ⊂ G and λ(G \ K) < �.
Proof. If A ∈ L0 , then there exists G ⊃ A open such that λ(G) <
λ∗ (A)+�/2 and there exists K ⊂ A compact such that λ(K) > λ∗ (A)−
�/2. Since K ⊂ G, we can write G = K ∪ (G \ K) as a disjoint union
of sets in L0 , and so λ(G) = λ(K) + λ(G \ K). That is,
λ(G \ K) = λ(G) − λ(K) < λ(A) + �/2 − (λ(A) − �/2) = �.
Date: September 25, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 7
For the other direction, ﬁx � > 0 and choose K ⊂ A ⊂ G such that
λ(G \ K) < �. Then
λ∗ (A) ≤ λ(G) = λ(K) + λ(G \ K) ≤ λ(K) + � ≤ λ∗ (A) + �.
Since this holds for any � > 0, we have λ∗ (A) ≤ λ∗ (A) ≤ λ∗ (A), and
�
hence λ∗ (A) = λ∗ (A).
Corollary 0.3. If A, B ∈ L0 , then A ∪ B, A ∩ B, and A \ B are all in
L0 .
Proof. By the approximation theorem, for any � > 0, we can ﬁnd K1 ⊂
A ⊂ G1 and K2 ⊂ B ⊂ G2 such hat λ(G1 \K1 ) < �/2 and λ(G2 \K2 ) <
�/2. Then K1 ∪ K2 ⊂ A ∪ B ⊂ G1 ∪ G2 , and so
(G1 ∪ G2 ) \ (K1 ∪ K2 ) = (G1 ∪ G2 ) ∩ (K1 ∪ K2 )c
= G1 ∩ (K1 ∪ K2 )c ∪ G2 ∩ (K1 ∪ K2 )c
⊂ G1 ∩ K1c ∪ G2 ∩ K2c
= (G1 \ K1 ) ∪ (G2 \ K2 ).
Thus,
λ(G1 ∪ G2 \ (K1 ∪ K2 )) ≤ λ(G1 \ K1 ) + λ(G2 \ K2 )
< �/2 + �/2 = �,
and A ∪ B ∈ L0 . Let Ki , Gi (i = 1, 2) be as before. Then K1 ∩ K2 ⊂
A ∩ B ⊂ G1 ∩ G2 . We have
(G1 ∩ G2 ) \ (K1 ∩ K2 ) = (G1 ∩ G2 ) ∩ (K1 ∩ K2 )c
= (G1 ∩ G2 ) ∩ (K1c ∪ K2c )
= (G1 ∩ G2 ∩ K1c ) ∪ (G1 ∩ G2 ∩ K2c )
⊂ (G1 ∩ K1c ) ∪ (G2 ∩ K2c )
⊂ (G1 \ K1 ) ∪ (G2 \ K2 ).
Thus,
λ(G1 ∩ G2 \ (K1 ∩ K2 )) < λ(G1 \ K1 ) + λ(G2 \ K2 )
< �/2 + �/2 = �.
The proof for A \ B is similar.
�
Countable additivity. Let Ak ∈ L0 for k = 1, 2, . . .. Let �
A = ∪∞
k=1 Ak
∞
∗
and assume λ (A) < ∞. Then A ∈ L0 and λ(A)
k=1 λ(Ak ).
� ≤
Furthermore, if the Ak ’s are disjoint, then λ(A) = ∞
λ(A
k ).
k=1
MEASURE AND INTEGRATION: LECTURE 7
3
Proof. First, the disjoint case. We have
∞
�
∗
λ (A) ≤
λ∗ (Ak ) (outer measure subadditivity)
=
k=1
∞
�
λ∗ (Ak )
k=1
≤ λ∗
�∞
(each Ak ∈ L0 )
�
Ak
≤ λ∗ (A).
k=1
�
Since λ∗ (A) = λ (A), A ∈ L0 , and it also follows that λ(A) = ∞
k=1 λ(Ak ).
In general, rewrite A as a disjoint union as follows. Let B1 = A1 ,
B2 = A2 \ A1 , B3 = A3 \ (A1 ∪ A2 ), and so on. Each Bk ∈ L0 , clearly
the Bk ’s are disjoint. It is straightforward to check that A = ∪∞
k=1 Bk :
the fact that the union is a subset of A is obvious, and if x ∈ Ak , then
x ∈ B1 or B2 or . . . Bk . From the preceding disjoint case we know that
∪∞
k=1 Bk ∈ L0 , and
�∞
�
∞
∞
�
�
λ(A) = λ
Bk =
λ(Bk ) ≤
λ(Ak ),
∗
k=1
k=1
k=1
where in the last step we noticed that each Bk ⊂ Ak .
�
Extension to arbitrary measurable sets. Let A ⊂ Rn . Then A is
Lebesgue measurable (and we write A ∈ L) if for all M ∈ L0 , we have
A ∩ M ∈ L0 . In this case, deﬁne
λ(A) = sup {λ(A ∩ M )}.
M ∈L0
Proposition 0.4. The new λ is consistent with all λ when λ∗ < ∞.
In other words, if A ⊂ Rn and λ(A) < ∞, then A ∈ L0 ⇐⇒ A ∈ L,
and the deﬁnitions of λ(A) agree.
Proof. If A ∈ L0 , then the lemma implies that A ∩ M ∈ L0 for all
M ∈ L0 ; thus A ∈ L. We know B(0, k), the ball of radius k with
center at the origin, is in L0 . Let Ak = A ∩ B(0, k). Then by deﬁnition
of L, we have Ak ∈ L. Also, A = ∪∞
k=1 Ak ∈ L0 by the countable
Next, take A ∈ L0 . Let λ̃(A) to be the new deﬁnition, that is,
λ̃(A) = sup {λ(A ∩ M )}.
M ∈L0
˜
Then A ∩ M ⊂⇒ λ(A ∩ M ) ≤ λ(A) ⇒ λ(A)
≤ λ(A). Since A ∈ L0 ,
˜
˜
choose M = A in deﬁnition of λ. Then λ(A) ≥ λ(A), and thus equality
must hold.
�
4
MEASURE AND INTEGRATION: LECTURE 7
Properties of (arbitrary) Lebesgue measurable sets.
(1) A ∈ L ⇒ Ac ∈ L0
(2) If Ai ∈ L (i = 1, 2, . . .), then ∪∞
Ai ∈ L .
i=1�
∞
(3) If Ai disjoint, then λ (∪∞
A
)
=
i
i=1
i=1 λ(Ai ).
(1) For M ∈ L0 , NTS that Ac ∩ M ∈ L0 . We know A ∩ M ∈
L0 . Since Ac ∩ M = M \ (A ∩ M ), and both M ∈ L0 and
A ∩ M ∈ L0 , we are done.
∞
(2) For M ∈ L0 , Ai ∩ M ∈ L0 . We have (∪∞
i=1 Ai ) ∩ M = ∪i=1 (Ai ∩
M ) and by countable additivity the last term is in L0 .
(3) Let A = ∪∞
M = ∪∞
is a disjoint
k=1 Ak . Then A ∩ �
k=1 (Ak ∩ M ) �
∞
∞
union. Thus, λ(A ∩ M ) =
k ∩ M) ≤
k=1 λ(A
k=1 λ(Ak ).
�∞
Taking sup over all M gives λ(A) ≤ k=1 λ(Ak ). For the other
direction, ﬁx N . Let M1 , . . . , MN ∈ L0 be arbitrary and put
M = ∪N
k=1 Mk . Then
�∞
�
λ(A) ≥ λ(A ∩ M ) = λ
(Ak ∩ M )
Proof.
=
∞
�
k=1
λ(Ak ∩ M )
k=1
≥
N
�
λ(Ak ∩ M )
k=1
≥
N
�
λ(Ak ∩ Mk ).
k=1
Since Mk are arbitrary, taking sup over all Mk gives λ(A) ≥
�N
�∞
k=1 λ(Ak ). Letting N → ∞, λ(A) ≥
k=1 λ(Ak ).
�
Corollary 0.5. By (1) and (2), L is a σ­algebra.
Corollary 0.6. By (3), λ is a positive measure on L, and thus (Rn , L, λ)
is a measure space.
MEASURE AND INTEGRATION: LECTURE 8
More properties of L.
(1) All open sets and closed sets are in L. (In particular, L contains
the Borel σ­algebra B.)
(2) If λ∗ (A) = 0, then A is measurable and λ(A) = 0. (All sets of
measure zero are measurable.)
(3) Approximation property: A ⊂ Rn is measurable ⇐⇒ for all
� > 0 there exists F ⊂ A ⊂ G with F closed, G open, and
λ(G \ F ) < �.
Proof.
(1) G open ⇒ G ∩ B(0, k) is measurable and open with
λ∗ < ∞. But G = ∪∞
k=1 G ∩ B(0, k) ∈ L since L is a σ­algebra.
Moreover, again using that L is a σ­algebra, all closed sets are
in L.
(2) We have 0 ≤ λ∗ (A) ≤ λ∗ (A) = 0 ⇐⇒ λ∗ (A) = λ∗ (A), so
A ∈ L0 and λ(A) = 0.
(3) First, assume the approximation property. Thus, for each k =
1, 2, . . ., there exists Fk ⊂ A ⊂ Gk , Fk closed, Gk open, such
that λ(Gk \ Fk ) < 1/k. Let B = ∪∞
k=1 Fk . By (1) and the fact
that L is a σ­algebra, B ∈ L. Also, B ⊂ A and A \ B ⊂
Gk \ B ⊂ Gk \ Fk . Thus,
λ∗ (A \ B) ≤ λ(Gk \ Fk ) < 1/k.
Since this holds for any k, λ∗ (A \ B) = 0. Thus, A \ B ∈ L and
λ(A \ B) = 0. But A = (A \ B) ∪ B, so A ∈ L.
For the converse, assume A ∈ L and let � > 0 be given. Let
Ek = B(0, k) \ B(0, k − 1) = {x ∈ Rn | k − 1 ≤ �x� < k}. Then
Ek ∈ L0 , so A∩Ek ∈ L0 . By the approximation property for L0 ,
there exist Kk ⊂ A ∩ Ek ⊂ Gk such that λ(Gk \ Kk ) < �/2k . Let
∞
F = ∪∞
k=1 Kk and G = ∪k=1 Gk . Since arbitrary unions of open
sets are open, G is open. Though this is not true for arbitrary
unions of closed sets, F is nevertheless closed. (Proof: Let x be
a limit point of F . Then x has to be a limit point of some Kk ,
Date: September 30, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 8
and since each Kk is closed, x ∈ Kk .) Now F ⊂ A ⊂ G and
�∞
�
∞
∞
G\F =
Gk \ F =
(Gk \ F ) ⊂
(Gk \ Fk ).
k=1
k=1
k=1
Hence,
λ(G \ F ) ≤
∞
�
λ(Gk \ Kk ) < �
k=1
∞
�
2−k = �.
k=1
�
(4) If A ∈ L and λ∗ (A) < ∞, we have that λ∗ (A) = λ∗ (A) = λ(A).
We claim this is true even if λ∗ (A) = ∞. If λ(A) < ∞, then by
the approximation property there exist F ⊂ A ⊂ G such that
λ(G \ F ) < 1. Then
λ(G) = λ(G \ A) + λ(A) ≤ λ(G \ F ) + λ(A) < 1 + λ(A) < ∞.
This is a contradiction; it must be that λ(A) = ∞.
Now consider
A ∩ B(0, 1) ⊂ A ∩ B(0, 2) ⊂ A ∩ B(0, 3) ⊂ · · · .
Then
λ(A) = lim λ(A ∩ B(0, k)) = ∞.
k→∞
Since A ∩ B(0, k) ∈ L0 for each k,
λ(A ∩ B(0, k)) = λ∗ (A ∩ B(0, k)) ≤ λ∗ (A),
and so λ∗ (A) = ∞.
(6) If A ⊂ B and B is measurable, then λ∗ (A) + λ∗ (B \ A) = λ(B).
Let G be an open set such that G ⊃ A. Then
λ(G) + λ∗ (B \ A) ≥ λ(B ∩ G) + λ∗ (B \ A)
≥ λ(B ∩ G) + λ∗ (B \ G)
= λ(B ∩ G) + λ(B \ G) = λ(B).
Since G is arbitrary, λ∗ (A) + λ∗ (B \ A) ≥ λ(B). Next take
K ⊂ B \ A compact. Then
λ∗ (A) + λ(K) ≤ λ∗ (B \ K) + λ(K)
= λ(B \ K) + λ(K) = λ(B).
Since K is arbitrary, λ∗ (A) + λ∗ (B \ A) ≤ λ(B).
MEASURE AND INTEGRATION: LECTURE 8
3
(7) (Carathéodory condition) A set A is measurable if and only if
for every set E ⊂ Rn ,
λ∗ (E) = λ∗ (E ∩ A) + λ∗ (E ∩ Ac ).
If A ∈ L, then let G ⊃ E be open. Then
λ(G) = λ(G ∩ A) + λ(G ∩ Ac ) ≥ λ∗ (E ∩ A) + λ∗ (E ∩ Ac ).
Since G is arbitrary,
λ∗ (E) ≥ λ∗ (E ∩ A) + λ∗ (E ∩ Ac ),
but
λ∗ (E) ≤ λ∗ (E ∩ A) + λ∗ (E ∩ Ac )
Conversely, let E, M ∈ L0 . We have assumed that λ∗ (M ) =
∗
λ (M ∩ A)λ∗ (M ∩ Ac ). From (6), since M ∩ Ac ⊂ M ,
λ(M ) = λ∗ (M ∩ Ac ) + λ∗ (M \ (M ∩ Ac ))
= λ∗ (M ∩ Ac ) + λ∗ (M ∩ A).
Thus, λ∗ (M ∩ A) = λ∗ (M ∩ A), and so A ∩ M ∈ L0 . Since M
is arbitrary, A ∈ L.
Discussion. The Carathéodory condition is quite signiﬁcant. It shows
that the knowledge of the properties of outer measure alone is suﬃcient
to decide which sets are measurable. Although this means that the
Lebesgue measure could have been developed entirely in terms of λ(I)
for special rectangles, our method of development is preferable for the
beginner.
MEASURE AND INTEGRATION: LECTURE 9
Invariance of Lebesgue measure. Given A ⊂ Rn and z ∈ Rn , let
z + A = {z + x | x ∈ A} be the translate of A by z. Given t > 0, let
tA = {tx | x ∈ A} be the dilation of A by t.
Let I = [a1 , b1 ] × · · · × [an , bn ] and z = z1 × · · · × zn . Then
z + I = [z1 + a1 , z1 + b1 ] × · · · × [zn + an , zn + bn ],
and
tI = [ta1 , tb1 ] × · · · × [tan , tbn ],
and we have
λ(z + I) = (z1 + b1 − z1 − a1 ) · · · (zn + bn − zn − az )
= (b1 − a1 ) · · · (bn − an )
= λ(I).
and
λ(tI) = tn · λ(I).
If P is a special polygon, then λ(z + P ) = λ(P ) and λ(tP ) = tn P .
�
Indeed, write P = N
i=1 Ii and the proof is straightforward.
If G is an open set, then λ(z + G) = λ(G) and λ(tG) = tn λ(G).
We have λ(G) = sup{λ(P ) | P ⊂ G special polygon}, so λ(z + G) =
sup{λ(P ) | P ⊂ z + G, P special polygon}. But P ⊂ G special polygon
⇐⇒ z + P ⊂ z + G special polygon. Since Lebesgue invariance holds
for special polygons, it holds for open sets.
Finally, by similar reasoning, it can be shown that a set A ⊂ Rn
is measurable if and only if z + A is measurable if and only if tA is
measurable, and λ(A) = λ(z + A), λ(tA) = tn λ(A).
A non­measurable set E ⊂ Rn . Let Q be the set of rational num­
bers. For x ∈ R, consider x+Q = {x+q | q ∈ Q}. Then y ∈ x+Q ⇐⇒
y − x ∈ Q.
Claim: if x, x� ∈ R, then either (i) x + Q = x� + Q or (ii) (x + Q) ∩
�
(x + Q) = ∅. Proof: If the intersection is nonempty, then there exists
y = x + q1 = x� + q2 , which implies that x − x� = q1 − q2 ∈ Q. Thus,
x + Q = x� + Q, and the claim is proved.
Date: October 2, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 9
We have shown that R is covered disjointly by the sets x + Q.
The Axiom of Choice states that there exists a set E ⊂ R such that
every point of R belongs to only one of these sets, i.e.,
R=
(x + Q)
x∈E
is a disjoint union. Alternatively, for any x ∈ R, there exists a unique
y ∈ E and unique z ∈ Q such that x = y + z.
Since the set Q is countable, its elements can be enumerated: Q =
{q1 , q2 , . . .}. Thus,
∞
R=
(qk + E)
k=1
is a disjoint union. Using outer measure subadditivity and invariance
of Lebesgue measure,
∗
λ (R) ≤
∞
�
∗
λ (qk + E) =
k=1
∞
�
λ∗ (E).
k=1
∗
Hence we must have that λ (E) > 0 (otherwise λ∗ (R) = 0).
Now let K ⊂ E be an arbitrary compact subset of E and let D =
(0, 1) ∩ Q. (The set D is a bounded countably inﬁnite set.) Then
(q + K) = D + K
q∈D
is a bounded set. The sets in the union are disjoint, since rational
translates of E are disjoint. We have
∞ > λ(D + K) (bounded)
�
�
=λ
(q + K)
=
�
q∈D
λ(q + K)
q∈D
=
�
λ(K).
q∈D
Since the sum is over an inﬁnite index set, λ(K) = 0. Because K ⊂ E
arbitrary ⇒ λ(K) = 0, we have λ∗ (E) = 0. But 0 = λ( E) < λ∗ (E) ⇒
E �∈ L.
Corollary 0.1. If A ⊂ Rn is measurable with positive measure, then
there exists B ⊂ A that is not measurable.
MEASURE AND INTEGRATION: LECTURE 9
3
Proof. Write A = ∪∞
k=1 ((qk + E) ∩ A) as a disjoint union. Then
0 < λ(A) = λ∗ (A) ≤
∞
�
λ∗ ((qk + E) ∩ A) ,
k=1
∗
and so λ ((qk + E) ∩ A) > 0 for some k. But λ∗ ((qk + E) ∩ A) ≤
λ∗ (qk + E) = λ∗ (E) = 0, a contradiction.
�
Invariance under linear transformations.
Theorem 0.2. Let T : Rn → Rn be a linear map and A ⊂ Rn . Then
λ∗ (T A) = |det T | λ∗ (A),
λ∗ (T A) = |det T | λ∗ (A).
If A is measurable, then T A is measurable and
λ(T A) = |det T | λ(A).
Proof. First assume that T is invertible, i.e., that det T �= 0. We will
use the following lemma.
Lemma 0.3. Let T be invertible and let J = [0, 1)n . Let ρ be deﬁned
by λ(T J) = ρλ(J). If A ⊂ Rn , then λ∗ (T A) = ρλ∗ (A) and λ∗ (T A) =
ρλ∗ (A). If A is measurable, then T A is measurable and λ(T A) =
ρλ(A).
Proof. The set J is the union of countably many compact sets:
∞
J=
[0, 1 − 1/k]n ,
k=1
and so
TJ =
∞
T ([0, 1 − 1/k]).
k=1
Since T maps compact sets to compact sets, T J is the union of count­
ably many compact sets. Thus, T J is measurable, so the deﬁnition of
ρ makes sense.
We just to need to prove that λ(T G) = ρλ(G) for G open. As before,
if the measure of open sets is invariant, then outer measure, compacts,
and inner measures are invariant.
Let G ⊂ Rn be open. Claim: can write G = ∪∞
k=1 Jk with Jk ’s disjoint
and each Jk is a translation and dilation of J. (Pair by integer of those
not contained, then pair by 1/2, then by 1/4, . . ..) Let Jk = zk + tk · J.
Then λ(Jk ) = tnk λ(J).
T Jk = T zk + tk · T J
4
MEASURE AND INTEGRATION: LECTURE 9
⇒ λ(T Jk ) = tnk λ(T J)
= tnk ρλ(J)
= tnk ρt1−n
λ(Jk ).
k
∞
Thus, λ(T Jk ) = ρλ(Jk ). Since G = ∪k=1 Jk , T G = ∪∞
k=1 T Jk , which is
a disjoint collection of measurable sets. Thus we have
∞
∞
�
�
λ(T G) =
λ(T Jk ) =
ρ · λ(Jk ) = ρ · λ(G).
k=1
k=1
�
To identify ρ, check for elementary matrices just on the cube. This
shows that in fact ρ = |det T |.
Lastly, if T is not invertible, i.e., det T = 0, then the image T Rn is
the subset of a hyperplane. This means that T A has measure zero, so
the formula still holds.
�
A linear transformation is a rotation when the matrix is an orthogo­
nal matrix: AAT = I. In this case, it must be that det A = ±1. Thus,
Lebesgue measure is invariant under rotation.
Finally, there is an important subgroup of the group of all n × n real
matrices known as the special linear group, denoted
SL(n, R) = {A | det A = 1}.
MEASURE AND INTEGRATION: LECTURE 10
Integration as a linear functional. A complex vector space is a set
V with two operations: addition (+) and scalar multiplication (·).
Addition: For all x, y, z ∈ V ,
• x + y = y + x.
• x + (y + z) = (x + y) + z.
• ∃ unique vector 0 such that x + 0 = x for all x.
• ∃ (−x) such that x + (−x) = 0.
Multiplication: For all α, β ∈ C, x ∈ V ,
• 1 · x = x
• α · (β · x) = (αβ) · x
• α · (x + y) = α · x + α · y
• (α + β) · x = α · x + β · x.
A linear transformation is a map Λ : V1 → V2 from a vector space V1
to a vector space V2 such that Λ(αxβ y) = αΛx + βΛy. If V2 = C (or
R), then Λ is a linear functional.
Let (X, M, µ) be a measure space. Then
�
�
�
1
L (µ) = f : X → C |
|f | dµ < ∞, f measurable .
�
�
X
Let g : X → C be
Note that X : f �→ X f dµ is a linear functional.
�
a bounded measurable function. Then f �→ X f g dµ is also a linear
functional.
Special case: X = Rn . Let
C(Rn , R) = {f : Rn → R | f continuous}.
The Riemann integral is a positive linear functional since f ≥ 0 ⇒
Λf ≥ 0, where Λ is the Riemann integral.
Riesz theorem. Let X be a topological space and C(X) be the set
of functions from X to R. If Λ : C → R is a positive linear functional,
then there
exists a σ­algebra M and unique measure µ on X such that
�
Λf = X f dµ. Conversely, given a measure, then Λ is a positive linear
functional.
Date: October 7, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 10
Topology. Let X be a topological space. The space X is Hausdorﬀ
� q there exist neighborhoods U and
if for all p, q ∈ X such that p =
V such that p ∈ U , q ∈ V , and U ∩ V = ∅. The space X is locally
compact if for all p ∈ X there exists a neighborhood U of p such that
U (the closure of U ) is compact. (Inﬁnite dimensional spaces are not
locally compact.)
Let f : X → R. If {x | f (x) > α} is open for all α, then f is lower
semicontinuous. If {x | f (x) < α} is open for all α, then f is upper
semicontinuous. Examples: χU for U open is lower semicontinuous and
χF for F closed is upper semicontinuous.
�
The support of a function f is deﬁned as the set supp f = {x | f (x) =
0}. An important set is the set of all functions with compact support:
Cc (X) = {f : X → C | supp f is compact}.
Since supp fg ⊂ (supp f ) ∪ (supp g), Cc (X) is a vector space.
Notation: (1) K � f means that K is compact, f ∈ Cc (X), 0 ≤
f (x) ≤ 1 for all x ∈ X, and f (x) = 1 for all x ∈ K. (2) f � V
means that V is open, f ∈ Cc (X), 0 ≤ f (x) ≤ 1 for all x ∈ X, and
supp f ⊂ V .
Urysohn’s lemma. Let X be a locally compact Hausdorﬀ space, K ⊂
V , K compact, U open. Then there exists f ∈ Cc (X) such that K �
f �V.
A corollary to Urysohn’s lemma is the existence of partitions of unity.
Let V1 , . . . , Vn be open subsets of X (a locally compact Hausdorﬀ space)
and K compact such that K ⊂ V1 ∪· · ·∪Vn . Then there exists functions
hi � Vi such that h1 (x) + · · · + hn (x) = 1.
Riesz representation theorem (for positive linear functionals).
Theorem 0.1. Let X be a locally compact Hausdorﬀ space. Let
Λ : Cc (X) → C
be a positive linear functional (positive when restricted to f : X →
R≥0 ). Then there exists a σ­algebra M in X which contains all the
Borel sets and a unique positive measure µ on M such that
�
(a) Λf = X f dµ for all f ∈ Cc (X).
(b) µ(K) < ∞ for all compact sets K ⊂ X.
(c) If E ∈ M, then
µ(E) = inf{µ(V ) |E ⊂ V, V open}.
(d) If E is open or E ∈ M with µ(E) < ∞, then
µ(E) = sup{µ(K) | K ⊂ E, K compact}.
MEASURE AND INTEGRATION: LECTURE 10
3
(e) If E ∈ M, A ⊂ E, and µ(E) = 0, then A ∈ M.
Proof. (Outline) We must show uniqueness.
By (d), the measure of open sets determined by measure of compact
sets, and so by (c) the measure of any set in M is determined by the
measure of compact sets. Assume we have µ1 and µ2 which satisfy
the conditions of the theorem, and let K be compact. For any � > 0,
choose U open such that K ⊂ U and µ2 (U ) < µ2 (K) + �. By Urysohn’s
lemma, there exists f ∈ Cc (X) such that K � f � V . Then
�
�
µ1 (K) =
χK dµ1 ≤
f dµ1 = Λf
X
and
�
Λf =
X
X
�
f dµ2 ≤
X
χV dµ2 = µ2 (V ) < µ2 (K) + �.
Since this holds for any � > 0, µ1 (K) ≤ µ2 (K), and by reversing the
roles of µ1 and µ2 , we have µ1 (K) = µ2 (K).
Now let V ⊂ X be open and deﬁne µ(V ) = sup{Λf | f � V }. For
E ⊂ X, deﬁne µ(E) = inf{µ(V ) | E ⊂ V, V open} = λ∗ (E). (λ∗ will
not be countably additive on all sets, only on the σ­algebra.) Let MF
be the set of E ⊂ X such that
µ(E) = sup{µ(K) | K ⊂ E, K compact} and µ∗ (E) < ∞.
Finally, M is simply E ⊂ X such that E ∩ K ∈ MF for all K ∈
MF .
�
Properties.
�
(1) µ∗ is countably subadditive: µ(∪Ei ) ≤ �µ(Ei ).
(2) If Ei ∈ MF are disjoint, then µ(∪Ei ) = µ(Ei ).
(3) MF contains all open sets.
(4) (Approximation) If E ∈ MF and � > 0, then there exist K ⊂
E ⊂ V , K compact, V open, such that µ(V \ K) < �.
(5) M is a σ­algebra that contains the Borel σ­algebra B, and µ is
(6) If f ∈ Cc (X), then Λf = X f dµ.
�
Proof. Just NTS that Λf ≤ X f dµ for f real in Cc (X). Then
�
�
−Λf = Λ(−f ) ≤
(−f )dµ = − f dµ
X
X
�
⇒ Λf ≥
f dµ.
X
The complex case follows from the real case by complex linear­
ity. Let f ∈ Cc (X) and supp f = K compact. The continuous
4
MEASURE AND INTEGRATION: LECTURE 10
image of compact sets is compact ⇒ f (K) ⊂ [a, b]. Choose
� > 0 and choose yi (i = 0, 1, . . . , n) such that yi − yi=1 < � and
y0 < a < y1 · · · < yn = b (i.e., partition the range by �). Let
Ei = {x | yi=1 < f (x) ≤ yi } ∩ K.
Since f is continuous, f is Borel measurable and ∪ni=1 Ei = K
is a disjoint union. choose open sets Vi ⊃ Ei such that µ(Vi ) <
µ(Ei ) + �/n for each i = 1, . . . , n and f (x) < yi + � for all x ∈ Vi .
(The latter can be done by continuity of f .)
�
By partition of unity,
there
exists
h
�
V
such
that
i
i
i hi = 1
�
on K. Write f = i hi f . Then
�
�
µ(K) ≤ Λ(
hi ) =
Λhi ,
i
i
hi f ≤ (yi + �)hi , and yi − � < f (x) ∀x ∈ Ei .
Thus,
n
n
�
�
Λf =
Λ(hi f ) ≤
(yi + �)Λhi
i=1
=
≤
=
n
�
i=1
(|a| + yi + �)Λhi − |a|
i=1
n
�
(|a| + yi + �)(µ(Ei ) + �/n) − |a| µ(K)
i=1
n
�
(|a| + �)(µ(Ei ))
= |a| µ(K) +
n
�
n
�
(|a| + yi + �)�/n +
i=1
(yi − �)µ(Ei ) + 2�µ(K) + �/n
�i=1
≤
Λhi
i=1
i=1
=
n
�
X
n
�
yi µ(Ei )
i=1
n
�
(|a| + yi + �)
i=1
f dµ + �(constant).
�
Deﬁnitions. A measure space (X, M, µ) is called a Borel measure if
B ⊂ M. If µ(E) = inf{µ(V ) | E ⊂ V, V open} for all E ∈ M,
then µ is called outer regular. Similarly, if µ(E) = sup{µ(K) | K ⊂
E, K compact} for all E ∈ M, then µ is called inner regular. If µ is
both inner and outer regular, it is said to be regular.
MEASURE AND INTEGRATION: LECTURE 10
5
A space X is σ­compact if X = ∪∞
i=1 Ki where each Ki is compact.
It is σ­ﬁnite if X = ∪∞
E
where
µ(E
i ) < ∞ for each i.
i=1 i
Addition to Riesz. If X is locally compact, σ­compact, Hausdorﬀ
space then we also have:
(1) If E ∈ M and � > 0, then there exists F ⊂ E ⊂ V , F closed,
V open, such that µ(V \ F ) < �.
(2) For all E ∈ M there exists A ⊂ E ⊂ B such that A is Fσ , B is
Gδ , and µ(B \ A) = 0.
�
Application. Let X = Rk , Λ : Cc (X) → R given by Λf = X f , the
Riemann integral. Then Lebesgue measure is what you get from the
Riesz theorem.
MEASURE AND INTEGRATION: LECTURE 11
Principles of measure theory.
(1) Every measurable set is nearly a Borel set: A = Fσ ∪N = Gδ ∪N
(N is a null set: a set of measure zero).
(2) Every measurable set is nearly an open set: λ(U ) < λ(E) + �.
(3) Every measurable function is nearly continuous (Lusin’s theo­
rem).
(4) Every convergent sequence of measurable functions is nearly
uniformly convergent (Egoroﬀ’s theorem).
Lusin’s theorem.
Theorem 0.1. Let f : X → R (or C) be a measurable function on
a locally compact Hausdorﬀ space X. Let A ⊂ X, µ(A) < ∞, and
f (x) = 0 if x �∈ A. Given � > 0, there exists g ∈ Cc (X) such that
� g(x)}) < �.
µ({x | f (x) =
Proof. Assume that 0 ≤ f ≤ 1 and A compact. Recall that if f : X →
[0, ∞] is measurable, then there exist simple measurable functions si
such that (a) 0 ≤ s1 ≤ · · · ≤ f and (b) si → f as i → ∞. (Proof: Let
δ= 2−n and for t ≥ 0 deﬁne kn (t) such that kδn ≤ t < (k + 1)δn , and
deﬁne
�
kn (t)δn 0 ≤ t < n;
ϕn (t) =
n
n ≤ t ≤ ∞.
Then ϕn (t) ≤ t and ϕn (t) → t as n → ∞. The function ϕ ◦ f is simple
and ϕn ◦ f → f as n → ∞.)
Next, let t1 = s1 , . . . , tn = sn − sn−1 . Claim: sn − sn−1 takes only
values 0 and 2−n . Let Tn ⊂ A where Tn = {x | tn = 2−n }. Then
∞
�
f (x) = s1 + (s2 − s1 ) + · · · =
tn .
n=1
Since X is locally compact, we may choose A ⊂ V , V open, and V
compact. There exists Kn ⊂ Tn ⊂ Vn ⊂ V , Kn compact, Vn open, such
there exist function
that µ(Vn \ Kn ) < 2−n �. By Urysohn’s lemma,
�∞
hn such that Kn � hn � Vn . Deﬁne g(x) = n=1 2−n hn (x). Since
Date: October 9, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 11
this series converges uniformly on X, g is continuous and supp f ⊂ V .
f (x) except on
But 2−n hn (X) = tn except on Vn \ Kn . Thus, g(x) = �
∞
n
∪∞
V
\
K
,
and
µ
of
this
set
is
less
than
or
equal
to
n
n=1 n
n=1 �/2 = �.
Thus, we have proved the case where 0 ≤ f ≤ 1 and A is compact.
Thus, it is true when f is a bounded measurable function and A is
compact.
Now look at
f
f=
.
(sup f ) + 1
If A is not compact and µ(A) < ∞, then there exists K ⊂ A such that
µ(A \ K) < � for any �. Let g = 0 on A \ K. For f not bounded, let
Bn = {x | |f (x)| > n}. Then ∩n Bn = ∅, so µ(Bn ) → 0. Then f agrees
with (1 − χBn )f except on Bn , and we can let g = 0 on Bn .
�
Corollary 0.2. Let f : X → R, A ⊂ X, µ(A) < ∞, f (x) = 0 if
x �∈ A, and |f (x)| < M for some M < ∞. Then there exists a sequence
gn ∈ Cc (X) such that |gn (x)| < M and f (x) = limn→∞ gn (x) almost
everywhere.
Proof. By the theorem, for n > 0, there exists gn ∈ Cc (X) such that
−n
. Let En = {x | f �= gn }. Then
sup |gn | ≤ sup |f | < M�and µ(En ) < 2�
∞
∞
En is measurable and n=1 µ(En ) = n=1 2−n < ∞. Claim:
� almost all
x ∈ X lie in at most ﬁnitely many Ek . Proof: Let g(x) = ∞
k=1 χEk (x).
Then x is in inﬁnitely many Ek ⇐⇒ g(x) = ∞. We have
�
∞ �
�
g(x) dµ =
χEk (x) dµ < ∞
X
k=1
X
by monotone convergence. Thus, g(x) < ∞ almost everywhere, which
�
implies that lim gn = f a.e.
Vitali­Caratheodory theorem.
Theorem 0.3. Let f : X → R and f ∈ L1 (µ). Given � > 0, there exists
functions u, v : X → R such that�u ≤ f ≤ v, u is upper semicontinuous,
v is lower semicontinuous, and (v − u)dµ < �.
Proof. Assume f ≥ 0. Choose 0 ≤ s1 ≤ · · · ≤ f simple and measurable
sn − sn−1 . Then tn has only ﬁnitely
such that lim sn = f . �Let tn = �
∞
many values and f = ∞
t
=
k=1 n
k=1 ck χEk . We have
�
∞
�
ci µ(Ei ) < ∞.
f dµ =
X
i=1
MEASURE AND INTEGRATION: LECTURE 11
3
Choose Ki ⊂ Ei ⊂ Vi , Ki compact, Vi open, such that ci µ(Vi \ Ki ) <
2−i−1 �. Let
∞
N
�
�
v=
ci χVi , u =
ci χKi ,
i=1
i=1
�
where N is chosen so that ∞
c
µ(E
)
< �/2. Then v is lower semi­
i
i
N +1
continuous, u is upper semicontinuous, and u ≤ f ≤ v. Also,
N
�
v−u=
≤
≤
i=1
∞
�
ci (χVi − χKi ) +
ci (χVi − χKi ) +
i=1
∞
�
∞
�
ci χVi
N +1
∞
�
ci (χVi − χVi + χKi )
i=N +1
ci χKi ,
i=N +1
and so
�
X
(v − u)dµ ≤
∞
�
ci µ(Vi \ Ki ) +
i=1
∞
�
ci χEi
N +1
< �/2 + �/2 = �.
�
MEASURE AND INTEGRATION: LECTURE 12
Appoximation of measurable functions by continuous func­
tions. Recall Lusin’s theorem.Let f : X → C be measurable, A ⊂ X,
µ(A) < ∞, and f (x) = 0 if x �∈ A. Given � > 0, there exists g ∈ Cc (X)
� g(x)}) < � and
such that µ({x | f (x) =
sup |g(x)| ≤ sup |f (x)| .
x∈X
x∈X
A corollary with the same assumptions and f bounded (i.e., |f (x)| <
M ) is that there exists sequence gn ∈ Cc (X), |gn | < M such that
lim gn (x) = f (x) almost everywhere.
Convergence almost everywhere. Lebesgue’s dominated conver­
gence theorem (LDCT) in the case of almost everywhere.
Theorem 0.1. Let f1 , f2 , . . . : X → C be a sequence of measurable
functions deﬁned a.e. Let g : X → C be deﬁned almost everywhere
and g ∈ L1 (µ). Assume limk→∞ fk (x) exists for a.e. x ∈ X and
|fk (x)| ≤ |g(x)| for a.e. x ∈ X. Then
� �
�
�
lim fk dµ = lim
fk dµ.
X
k→∞
k→∞
X
Proof. Let Ek = {x | |fk (x)| ≥ |g(x)|}. Then µ(Ek ) = 0. Let E =
∪∞
k=1 Ek . Then µ(E) = 0. Redeﬁne fk = 0 on E; this does not change
the integrals. Now |fk | ≤ |g | a.e., and we can apply the regular LDCT.
�
Theorem �
0.2. Let
f1 , f2 , . . . : X → C�
with each fk ∈ L1 (µ) and as­
∞ �
sume that k=1 X |fk |dµ < ∞. Then ∞
k=1 fk exists a.e. and
�
�
�
∞
∞ �
�
�
fk dµ =
fk dµ.
X
k=1
k=1
X
�
�∞
Proof.
Let
g
=
|f
|
.
Monotone
convergence
implies
that
g =
k
� �∞
�∞k=1�
1
|
f
|
=
|f
|
<
∞.
Thus
g
∈
L
(µ)
and
so
g
<
∞
a.e.
k
k=1
k=1
�∞
�∞k
Thus, k=1 |fk (x)| < ∞ a.e. This implies that the series k=1 fk (x)
Date: October 14, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 12
�
converges absolutely a.e. Let Fj = jk=1 fk . Then Fj is dominated by
g for all j, and we can apply LDCT. We have
� ��
∞
X
�
fk
�
dµ =
�
lim Fj dµ = lim
X j→∞
k=1
= lim
� �
j
j→∞
∞ �
�
=
k=1
j→∞
X k=1
X
X
fk dµ = lim
j→∞
Fj dµ
j �
�
k=1
X
fk dµ
fk dµ.
�
Countable additivity of the integral. Let E1 , E2 , . . . be a countable
sequence of measurable sets. Let E = ∪∞
k=1 Ek and f : X
� → C be
1
measurable.
Assume
either
f
≥
0
or
f
∈
L
(E)
(i.e.,
f dµ =
E
�
f χE dµ < ∞). Then
X
�
f dµ =
E
∞ �
�
k=1
Ek
f dµ.
Proof. First let f ≥ 0. Then
�
E
�
f dµ =
=
∞ �
�
k=1
X
X
f χE dµ
� �
∞
E k=1
f χEk dµ
f χEk dµ
=
∞ �
�
k=1
Ek
f dµ.
Now let f ∈ L1 (E) and fk = χEk . By the previous theorem, we need
only check the convergence of the series of integrals of |f χEk |.
MEASURE AND INTEGRATION: LECTURE 12
We have
∞ �
�
k=1
X
|fk | dµ =
∞ �
�
k=1
=
X
3
|f | χEk dµ
∞ �
�
Ek
|f | dµ
�k=1
=
|f | dµ < ∞,
E
because of the case when f ≥ 0.
�
MEASURE AND INTEGRATION: LECTURE 13
Egoroﬀ ’s theorem (pointwise convergence is nearly uniform.
Theorem 0.1. Suppose µ(X) < ∞. Let fn : X → C be a sequence of
measurable functions such that fn → f a.e. For all � > 0, there exists
a measurable subset E ⊂ X with µ(X \ E) < � and such that fn → f
uniformly on E.
Proof. Let
S(n, k) =
�
{x | |fi (x) − fj (x)| < 1/k}.
i,j>n
Clearly, S(n, k) is measurable, since it is the countable intersection of
measurable sets. Note that
�
�
S(n, k) =
{· · · } ∩
{· · · }
i,j>n+1
i=n,j>n+1
= S(n + 1, k) ∩ {stuﬀ}.
Thus, S(n, k) ⊂ S(n + 1, k), that is, for each k, we have an ascending
sequence of sets. Claim: for each k, X = ∪∞
n=1 S(n, k). Given k,
and x ∈ X, we know fi (x) → f (x). Thus there exists N such that
|fi (x) − fj (x)| < 1/k for all i, j > N since any convergent sequence is
Cauchy. Thus, x ∈ S(N, k). Obviously
∞
X⊃
S(N, k) ⇒ X =
n=1
∞
S(n, k)
n=1
for each k. So we have
lim µ(S(n, k)) → µ(X)
n→∞
for any k.
For each k = 1, 2, . . ., choose nk so that
|µ(S(nk , k) − µ(X)| < �/2k .
Date: October 16, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 13
∞
(Recall µ(X) < ∞.) Let E = ∩∞
k=1 S(nk , k). Then X \ E = ∪k=1 (X \
S(nk , k)). Thus,
∞
�
µ(X \ E) ≤
µ(X \ S(nk , k))
=
k=1
∞
�
µ(X) − µ(S(nk , k))
k=1
≤
∞
�
|µ(X) − µ(S(nk , k))|
k=1
<
∞
�
�/2k = �.
k=1
Claim: fn → f uniformly on E; that is, given any δ > 0, there
exists N such that |fi (x) − fj (x)| < δ for all i, j > N and every
x ∈ E. Choose k such that 1/k < δ. If x ∈ S(n, k), by deﬁ­
nition |Fi (x) − fj (x)| < δ for all i, j > n. In particular, for x ∈
S(nk , k), |fi (x) − fj (x)| < δ for all i, j > nk . But S(nk , k) ⊃ E, so
|fi (x) − fj (x)| < δ for all i, j > nk and all x ∈ E.
�
The theorem is not necessarily true if µ(X) = ∞. For example, if µ
is Lebesgue measure on R and fn = χ[n,n+1] . Then fn → 0 pointwise,
but for any n �= m, |fn (x) − fm (x)| = 1 on a set of measure 2.
Convergence
in measure. Here is an example. Let fn : X → R and
�
|fn | → 0. Then if � > 0,
X
�
intX |fn | ≥
|fn | > �µ({x | fn (x) > �}).
{x | fn >�}
�
So, given � > 0, choose N such that for all n > N , X |fn | < �2 . Then
� ≥ µ({x | fn (x) > �} for all n > N .
We say that fn → f in measure if given � > 0 there exists N such
that, for all n ≥ N , µ({x | |f (x) − fn (x)| > �}) < �.
Convergence almost everywhere implies convergence in mea­
sure.
Theorem 0.2. If fn → f a.e. and µ(X) < ∞, then fn → f is
measure.
Proof. Let A = {x | fn (x) → f (x)}. Then µ(X \ A) = 0. Since
µ(X) < ∞, µ(A) < ∞ and we may apply Egoroﬀ’s theorem. Thus,
there exists a set E such that µ(A \ E) < � and fn → f uniformly on
E. Given � > 0, there exists N such that |f (x) − fn (x)| < � for all
MEASURE AND INTEGRATION: LECTURE 13
3
n > N and all x ∈ E. So, for n > N , |f (x) − fn (x)| can be greater
than � only on (A \ E) ∪ (X \ A). This means that
µ({x | |fn (x) − f (x)| > �}) ≤ µ(A \ E) + µ(X \ A)
<�+0=�
for all n > N .
�
However, if fn → f in measure, then it is not true
� 1 that fn → f a.e.
For example, fn : [0, 1] → [0, 1] such that limn→∞ 0 fn (x) dx = 0 but
fn (x) → 0 for no x.
Convergence in measure implies some subsequence conver­
gence almost everywhere.
Theorem 0.3. If fn → f in measure, then fn has a subsequence fnk
such that limk→∞ fnk = f a.e.
Proof. Let � = 2−k . Given k, there exists N (k) such� that for n ≥ N (k),
�
µ({x | |f (x) − fn (x)| > 2−k } < 2−k . Let Ek = {x | �fN (k) (x) − f (x)� >
c
∞
∞
c
2−k }. Then µ(Ek ) < 2−k . If x �∈ ∪∞
i=k Ei , then x ∈ (∪i=k Ei ) = ∩i=k Ei .
Then
�
�
�
fN (i) (x) − f (x)�
< 2−i for every i ≥ k
⇒ fN (i) (x) → f (x).
Let
∞ ∞
�
A=
Ei .
k=1 i=k
So if x �∈ A, then fN (i) (x) → f (x). For any k,
µ(A) ≤
µ (∪∞
i=k Ei )
≤
∞
�
2−i = 2−k+1 ,
i=k
so µ(A) = 0.
�
Dominated convergence theorem holds for convergence in mea­
sure. We know that dominated convergence and monotone conver­
gence still hold if we replace convergence with convergence almost ev­
erywhere. Now we show that the theorems are valid if we replace
convergence by convergence in measure.
Theorem 0.4. Let fn : X → C be a sequence of measurable functions
deﬁned a.e. Suppose fn → f in measure and |fn | ≤ |g| a.e. with
g ∈ L1 (µ). Then
�
X
f dµ = lim fn dµ.
n→∞
4
MEASURE AND INTEGRATION: LECTURE 13
Note that the pointwise limit of fn may not exist.
Proof. Take any subsequence fnk . Clearly, fnk → f in measure. There
exists a subsequence f(nk )� such that f(nk )� → f pointwise a.e. Apply
dominated convergence to this subsequence. Then
�
�
f dµ = lim
f(nk )� dµ.
X
�→∞
X
�
Lemma 0.5. Let an be a sequence. If every subsequence has a subse­
quence which converges to α, then limn→∞ an = α.
MEASURE AND INTEGRATION: LECTURE 14
Convex functions. Let ϕ : (a, b) → R, where −∞ ≤ a < b ≤ ∞.
Then ϕ is convex if ϕ((1 − t)x + ty) ≤ (1 − t)ϕ(x) + tϕ(y) for all
x, y ∈ (a, b) and t ∈ [0, 1]. Looking at the graph of ϕ, this means that
(t, ϕ(t)) lies below the line segment connecting (x, ϕ(x)) and (y, ϕ(y))
for x < t < y.
Convexity is equivalent to the following. For a < s < t < u < b,
ϕ(u) − ϕ(t)
ϕ(t) − ϕ(s)
≤
.
t−s
u−t
If ϕ is diﬀerentiable, then ϕ is convex on (a, b) if and only if, for a <
s < t < b, ϕ� (s) ≤ ϕ� (t). If ϕ is C 2 (continuously twice diﬀerentiable),
then ϕ� increasing ⇒ ϕ�� ≥ 0.
Theorem 0.1. If ϕ is convex on (a, b), then ϕ is continuous on (a, b).
Jensen’s inequality. Let (Ω, M, µ) be a measure space such that
µ(Ω) = 1 (i.e., µ is a probability measure). Let f : Ω → R and f ∈
L1 (µ). If a < f (x) < b for all x ∈ Ω and ϕ is convex on (a, b), then
��
� �
ϕ
f dµ ≤ (ϕ ◦ f )dµ.
Proof. Let t =
Ω
�
Ω
f dµ. Since a < f < b,
�
a = a · µ(Ω) <
f dµ < b · µ(Ω) = b,
Ω
Ω
so a < t < b. Conversely,
ϕ(u) − ϕ(t)
ϕ(t) − ϕ(s)
≤
.
t−s
u−t
Fix t, and let
ϕ(t) − ϕ(s)
B = sup
.
t−s
a<s<t
Then ϕ(t) − ϕ(s) ≤ B(t − s) for s < t. We have
B≤
Date: October 21, 2003.
ϕ(u) − ϕ(t)
u−t
1
2
MEASURE AND INTEGRATION: LECTURE 14
for any u ∈ (t, b), so B(u − t) ≤ ϕ(u) − ϕ(t) for u > t. Thus ϕ(s) ≥
ϕ(t) + B(s − t) for any a < s < b. Let s = f (x) for any x ∈ Ω. Then
ϕ(f (x)) − ϕ(t) − B(f (x) − t) ≥ 0
for all x ∈ Ω.
Now ϕ convex ⇒ ϕ continuous, so ϕ ◦ f is measurable. Thus, inte­
grating with respect to µ,
�
�
�
(ϕ ◦ f )dµ −
ϕ(t) dµ − B
f dµ ≥ 0,
X
X
X
and the inequality follows.
�
Examples.
(1) Let ϕ(x) = ex be a convex function. Then
��
� �
exp
f dµ ≤
ef dµ.
Ω
Ω
(2) Let Ω = {p1 , . . . , pn } be a ﬁnite set of points and deﬁne µ({pi }) =
1/n. Then µ(Ω) = 1. Let f : Ω → R with f (pi ) = xi . Then
�
Ω
=
Thus
�
exp
f dµ =
1
(x1 + · · · + xn ).
n
n
�
f (pi )µ({pi })
i=1
� �
1
(x1 + · · · + xn ) ≤
ef dµ
n
Ω
1 x1
≤ (e + · · · + exn ).
n
Let yi = exi . Then
(y1 + · · · + yn )1/n ≤
1
(y1 + · · · + yn ),
n
which is the inequality between arithmetic �
and geometric means.
We also could take µ({pi }) = αi > 0 and ni=1 αi = 1. Then
y1α1 y2α2 · · · ynαn ≤ α1 y1 + α2 y2 + · · · + αn yn .
MEASURE AND INTEGRATION: LECTURE 14
3
Hölder’s and Minkowski’s inequalities. We deﬁne numbers p and
q to be conjugate exponents if 1/p + 1/q = 1. The conjugate exponent
of 1 is ∞. Conjugate exponents are the same if and only if p = q = 2.
Theorem 0.2. Let p and q be conjugate exponents with 1 < p < ∞.
Let (X, M, µ) be a measure space and f, g : X → [0, ∞] measurable
functions. Then
��
�1/p ��
�1/q
�
p
q
f g dµ ≤
f dµ
g dµ
(Hölder’s)
X
and
��
X
X
X
�1/p ��
�1/p ��
�1/p
p
p
(f + g) dµ
≤
f dµ
+
g dµ
(Minkowski’s).
p
X
X
Proof. �H¨older’s. Without loss
� ofp generality� weq may assume that
q
� 0, then let
1 and X g = 1. Indeed, if f �= 0 and g =
f = ��
f
fp
X
�1/p ,
g = ��
g
fp
X
�
X
fp =
�1/p .
�
(Otherwise, if f p = 0, then f p = 0 a.e., and both sides of the inequal­
ity are equal to zero.) We claim that
(0.1)
1
1
ab ≤ ap + bq for all a, b ∈ [0, ∞].
p
q
It is easy to check if a or b equals 0 or ∞. Assume 0 < a < ∞ and
0 < b < ∞, and write a = es/p and b = et/q for some s, t ∈ R. Let
Ω = {x1 , x2 }, µ(x1 ) = 1/p, and µ(x2 ) = 1/q. We have
�
�
exp f dµ ≤
ef dµ,
Ω
Ω
where f (x1 ) = s and f (x2 ) = t. Thus,
�
�
1
1
s t
+
≤ es + et ,
exp
p q
q
p
so (0.1) follows. Thus,
�
�
�
1
1
1 1
p
a dµ +
bq dµ = + = 1.
(f g)dµ ≤
p X
q X
p q
X
Minkowski’s. Observe that
(f + g)p = f (f + g)p−1 + g(f + g)p−1 .
4
MEASURE AND INTEGRATION: LECTURE 14
Since p and q are conjugate exponents, q = p/(p − 1). Thus,
��
�1/p ��
�(p−1)/p
�
p−1
p
(p−1)p/(p−1)
f (f + g)
≤
f
(f + g)
��
=
Similarly,
�
f
p−1
f (f + g)
p
�1/p ��
�(p−1)/p
.
(f + g)
��
≤
p
f
p
�1/p ��
p
�(p−1)/p
(f + g)
.
Let Ω = {x1 , x2 }, µ(x1 ) = 1/2 = µ(x2 ), and ϕ = tp . Then
��
�p �
f dµ ≤
f p dµ,
Ω
so
�
Thus,
a+b
2
Ω
�p
≤
ap b p
+ .
2
2
�
�
�
1
1
1
p
p
f +
g p < ∞.
(f + g) ≤
2p
2
2
Since 1 − (p − 1)/p = 1/p,
��
�1/p ��
�1/p ��
�1/p
p
p
p
(f + g) dµ
≤
f dµ
+
g dµ
.
X
X
X
�
MEASURE AND INTEGRATION: LECTURE 15
Lp spaces. Let 0 < p < ∞ and let f : X → C be a measurable
function. We deﬁne the Lp norm to be
��
�1/p
p
�f �p =
|f | dµ
,
X
p
and the space L to be
Lp (µ) = {f : X → C | f is measurable and �f �p < ∞}.
Observe that �f �p = 0 if and only if f = 0 a.e. Thus, if we make
the equivalence relation f ∼ g ⇐⇒ f = g a.e, then �·� makes Lp a
normed space (we will deﬁne this later).
If µ is the counting measure on a countable set X, then
�
�
f dµ =
f (x).
X
p
x∈X
p
Then L is usually denoted � , the set of sequences sn such that
�∞
�1/p
�
|sn |p
< ∞.
n=1
A function f is essentially bounded if there exists 0 ≤ M < ∞ such
that |f (x)|
≤ M for a.e. x ∈ X. The space L∞ is deﬁned as
L∞ (µ) = {f : X → C | f essentially bounded}
with the L∞ norm
�f �∞ = inf{M | |f (x)| ≤ M a.e. x ∈ X}.
Proposition 0.1. If f ∈ L∞ , then |f (x)| ≤ �f �∞ a.e.
Proof. By deﬁnition of inf, there exists Mk → �f �∞ such that |f (x)| <
Mk a.e, or, equivalently, there exists Nk with µ(Nk ) = 0 such that
|f (x)| ≤ Mk for all x ∈ Nkc . Let N = ∪∞
k=1 Nk . Then µ(N ) = 0. If
c
x ∈ N c = ∩∞
(N
)
,
then
|
f
(x)|
≤
M
k
k since Mk → �f �∞ . Thus,
k=1
c
|f (x)| ≤ �f �∞ for all x ∈ N .
�
Date: October 23, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 15
Theorem 0.2. Let 1 ≤ p ≤ ∞ and 1/p + 1/q = 1. Let f ∈ Lp (µ) and
g ∈ Lq (µ). Then f g ∈ L1 (µ) and
�f g�1 ≤ �f �p �g�q i.e.,
��
�1/p ��
�1/q
�
p
q
|f g| dµ ≤
|f |
|g |
.
Proof. If 1 < p < ∞, this is simply Hölder’s inequality. If p = 1,
q = ∞, then |f (x)g(x)| ≤ �g�∞ |f (x)| a.e. Thus,
�
�
|f g| ≤ �g� |f | .
�
Theorem 0.3. Let 1 ≤ p ≤ ∞. Let f, g ∈ Lp (µ). Then f + g ∈ Lp (µ)
and �f + g�p ≤ �f �p + �g�p .
Proof.� If 1 < p <�∞, this� is simply Minkowski’s inequality. If p = 1,
then |f + g| ≤ |f | + |g | is true. If p = ∞, then |f + g | ≤ |f | +
|g | ⇒ �f + g�∞ ≤ �f �∞ + �g �∞ .
�
Normed space and Banach spaces. A normed space is a vector
space V together with a function �·� : V → R such that
(a) 0 ≤ �x� < ∞.
(b) �x� = 0 ⇐⇒ x = 0.
(c) �αx� = |α| �x� for all α ∈ C.
(d) �x + y� ≤ �x� + �y�.
For example, Lp (µ) is a normed space if two functions f, g are consid­
ered equal if and only if f = g a.e. Also, Rn with the Euclidean norm
is a normed space.
A metric space is a set M together with a function d : M × M → R
such that
(a) 0 ≤ d(x, y) < ∞.
(b) d(x, x) = 0.
� y.
(c) d(x, y) > 0 if x =
(d) d(x, y) = d(y, x).
(e) d(x, y) ≤ d(x, z) + d(z, y).
A normed space is a metric space with metric d(f, g) = �f − g�.
Recall that xi → x ∈ M if limn→∞ d(xn , x) = 0. A sequence {xi } is
Cauchy if for every � > 0 there exists N (�) such that d(xj , xk ) ≤ � for
all j, k ≥ N (�).
Claim: if xn → x, then it is Cauchy. We know that limn→∞ d(xn , x) =
0, so given � > 0, there exists N such that d(xk , x) < �/2 for all k > N .
for j, k > N , d(xk , xj ) ≤ d(xj , x) + d(x, xk ) < �.
MEASURE AND INTEGRATION: LECTURE 15
3
However, a Cauchy sequence does not have to converge. For example,
consider the space R \ {0} (the punctured real line) with the absolute
value norm. The sequence xn = 1/n is Cauchy but it does not converge
to a point in the space.
A metric space is called complete if every Cauchy sequence converges.
By the Bolzano­Weierstrass theorem, Rn is complete. (Every Cauchy
sequence is bounded, so it has a convergent subsequence and must
converge.)
A normed space (V, �·�) that is complete under the induced metric
d(f, g) = �f − g� is called a Banach space.
Riesz­Fischer theorem.
Lemma 0.4. If {fn } is Cauchy, then there exists a subsequence fnk
such that d(fnk+1 , fnk ) ≤ 2−k .
Theorem 0.5. For 1 ≤ p ≤ ∞ and for any measure space (X, M, µ),
the space Lp (µ) is a Banach space.
Proof. Let 1 ≤ p < ∞ and let {fn } ∈ Lp (µ) be a Cauchy sequence.
By the lemma,
there� exists a subsequence
�
� nk� with n1 <� n2 < · · ·
such that �
fnk+1 , fnk �p < 2−k . Let gk = ki=1 �fni+1 − fni �p and g =
�
�
�
�fn − fn �
. By Minkowski’s inequality,
limk→∞ gk = ∞
i+1
i
i=1
p
k
k
�
�
�
�
�
�
�gk �p ≤
fni+1 − fni p <
2−i < 1.
i=1
i=1
Consider gkp . By Fatou’s lemma,
�
�
p
lim inf gk ≤ lim inf gkp ,
and so
�
g p ≤ 1 ⇒ g(x) < ∞ a.e.
Thus, the series
fn1 (x) +
∞
�
(fni+1 (x) − fni (x))
i=1
converges absolutely a.e. Deﬁne
�
�
fn1 (x) + ∞
i=1 (fni+1 (x) − fni (x)) where it converges;
f (x) =
0
otherwise.
4
MEASURE AND INTEGRATION: LECTURE 15
The partial sum
fn1 (x) +
k−1
�
(fni+1 (x) − fni (x)) = fnk (x),
i=1
and so
lim fnk (x) = f (x) a.e.
k→∞
Thus we have shown that every Cauchy sequence has a convergent
subsequence, and we NTS that fnk → f in Lp .
Given � > 0, there exists N such that �fn − fm �p < � for all n, m >
N . We have that
|f − fm |p = lim inf |fnk − fm |p
since fnk → f a.e. Thus,
�
�
p
|f − fm | =
lim inf |fnk − fm |p
X
X
�
≤ lim inf
|fnk − fm |p
< �p .
X
This implies that �f − fm �p < �, and thus
�f �p = �f − fm + f + m�p ≤ �f − fm �p + �fm �p < ∞.
We conclude that f ∈ Lp and �f − fm �p → 0 as m → ∞.
Now let p = ∞ and let {fn } be a Cauchy sequence in L∞ (µ). Let
Ak = {x | |fk (x)| > �fk �∞ }
and
Bm,n = {x | |fn (x) − fm (x)| > �fn − fm �∞ }.
These sets all have measure zero. Let
�
�∞
� � ∞
Bm,n .
N=
Ak ∪
k=1
n,m=1
Then N has measure zero.
For x ∈ N c , fn is a Cauchy sequence of complex numbers. Thus,
fn → f by completeness of C uniformly. Since �fk �∞ is bounded,
|fk (x)| < M for all x ∈ N c . Thus, f (x) < M for all x ∈ N c . Letting
f = 0 on N , we have �f �∞ < ∞ and �fn − f �∞ → 0 as n → ∞.
�
Theorem 0.6. Let 1 ≤ p ≤ ∞ and {fn } be a Cauchy sequence in Lp (µ)
such that �f − fn �p → 0. Then fn has a subsequence which converges
pointwise almost everywhere to f (x).
MEASURE AND INTEGRATION: LECTURE 15
5
Proof. Since �f − fn �p → 0, fn → f in measure. By the previous
theorem, there exists a subsequence which converges a.e.
�
Examples in R.
(1) A sequence in Lp can converge a.e. without converging in Lp .
Let fk = k 2 χ(0,1/k) . Then
��
�1/p
2p
�fk �p =
k
= k 2 (1/k)1/p = k 2−1/p < ∞.
(0,1/k)
p
Thus fk ∈ L and fk → 0 on R, but �fk �p → ∞.
(2) A sequence can converge in Lp without converging a.e. (HW
problem).
(3) A sequence can belong to Lp1 ∩ Lp2 and converge in Lp1 without
converging in Lp2 . Let fk = k −1 χ(k,2k) . Then fk → 0 pointwise
and �fk �p = k −1 k 1/p = k 1/p−1 . If p > 1, then �fk �p → 0 as
k → ∞, so fk → 0 in Lp norm. But �fk �1 = 1 so fk �→ 0 in L1 .
MEASURE AND INTEGRATION: LECTURE 16
Cc dense in Lp for 1 ≤ p < ∞.
Theorem 0.1. Let
S = {s : X → C | s simple, measurable such that µ({x | s(x) =
� 0})} .
For 1 ≤ p < ∞, S is dense in Lp (µ), i.e., given f ∈ Lp (µ) there exists
sequence sk ∈ S such that �sk − f �p → 0.
Proof. Note that S ⊂ Lp (µ) since
�
� 0}) < ∞.
sp dµ ≤ max sp µ({x | f (x) =
X
If f : X → R and f ≥ 0, then by the approximation theorem, there
exists sk simple measurable functions
�
� such that 0 ≤ s1 ≤ · · · ≤ f and
limk→∞ sk = f . Since sk ≤ f , spk ≤ f p < ∞. Thus,
sk ∈ Lp ⇒ sk ∈ S.
We have
f − sn ≤⇒ |f − sn |p ≤ |f |p .
So |f − sn |p ≤ f = |f |p ∈ L1 , and we can apply the dominated conver­
gence theorem. Thus,
�
�
p
lim
|f − sn | dµ =
lim(f − sn )p dµ − 0,
X
X
and so �sn − f �p → 0. If f is not non­negative, apply separately to f +
and f − .
�
Corollary 0.2. If X is a locally compact Hausdorﬀ space, then for
1 ≤ p < ∞, Cc (X) is dense in Lp .
Proof. Let S be as in the previous theorem. If s ∈ S and � > 0,
there exists g ∈ Cc (X) such that µ({x | g(x) =
� s(x)}) < � by Lusin’s
Date: October 28, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 16
theorem, and also |g| ≤ �s�∞ . Thus,
��
�1/p ��
p
�g − s�p =
=
|g − s|
��
=
�
g=s
p
g=s
�1/p
|g − s|
p
|g − s| +
��
≤
�
�
g=s
p
2
�s�p∞
g=s
�
|g − s|
p
�1/p
�1/p
≤ 2 �s�∞ �1/p .
Thus, Cc (X) is dense in S, and since S is dense in Lp , Cc (X) is dense
�
in Lp .
Here is �an example. Let X = Rn and let f, g ∈ Cc (Rn ). Deﬁne
∞
d(f, g) = −∞ |f (t) − g(t)| dt. Note that Cc (Rn ) ⊂ L1 (Rn ) and L1 is
complete. The space L1 (Rn ) is the completion of Cc (Rn ) under this
metric, provided f ∼ g if f = g a.e. Any metric space has a unique
completion under its metric.
The case p = ∞. Let f, g ∈ Cc (X) and
d(f, g) = sup |f (x) − g(x)| .
x∈X
∞
Then L is not the completion of Cc (X) under d.
A function f : X → C vanishes at inﬁnity if for every � > 0 there
exists a compact subset K ⊂ X such that |f (x)| < � whenever x �∈ K.
The set of all continuous function that vanish at inﬁnity is denoted by
C0 (x).
Cc dense in C0 .
Theorem 0.3. The completion of Cc (X) under �·�∞ is C0 (X).
Proof. We show that (a) Cc (X) is dense in C0 (X), and (b) C0 (X) is
complete.
Proof of (a). Let f ∈ C0 (X). For � > 0, there exists K compact
such that |f (x)| < � for all x ∈ K c . By Urysohn’s lemma, there exists
g ∈ Cc (X) such that K � g, 0 ≤ g ≤ 1, and g = 1 on K. Let h = f g.
Then h ∈ Cc (X) and �f − h�∞ < �. (f = h on K and f < � on K c .)
Proof of (b). Let fn be a Cauchy sequence in C0 (X), i.e., given
� > 0, there exists N such that i, j > N , �fi (x) − fj (x)�∞ < �. In
other words, fn converges uniformly. Thus,
lim fn = f exists
n→∞
and f is continuous. Given � > 0, there exists n such that �fn − f �∞ <
�/2 and there exists K compact such that |fn (x)| < �/2 for all x ∈ K c .
Then |f | = |f − fn + fn | ≤ �/2 + �/2 = � on K c . Thus, f ∈ C0 (X). �
MEASURE AND INTEGRATION: LECTURE 17
Inclusions between Lp spaces. Consider Lebesgue measure on the
space (0, ∞) ⊂ R. Recall that xa is integrable on (0, 1) ⇐⇒ a > −1,
and it is integrable on (1, ∞) ⇐⇒ a < −1. Now let 1 ≤ p < q ≤ ∞.
Choose b such that 1/q < b < 1/p. Then x−b χ(0,1) is in Lp but not in
Lq , which shows that Lp �⊂ Lq . On the other hand, x−b χ(1,∞) is in Lq
but not in Lp , so that Lq �⊂ Lp . Thus, in general there is no inclusion
relation between two Lp spaces.
The limit of �f �p as p → ∞. For convenience, deﬁne �f �p to be ∞
if f is M­measurable but f �∈ Lp .
Theorem 0.1. Let f ∈ Lr for some r < ∞. Then
lim �f �p
= �f �∞ .
p→∞
This justiﬁes the notation for the L∞ norm.
Proof. Let t ∈ [0, �f �∞ ). By deﬁnition, the set
A = {x ∈ X | |f (x)| ≥ t}
has positive measure. Observe the trivial inequality
��
�1/p
p
�f �p ≥
|f | dµ
A
≥ (t µ(A))1/p
p
= tµ(A)1/p .
If µ(A) is ﬁnite, then µ(A)1/p → 1 as p → ∞. If µ(A) = ∞, then
µ(A)1/p = ∞. In both cases, we have
lim inf �f �p ≥ t.
p→∞
Since t is arbitrary,
lim inf �f �p ≥ �f �∞ .
p→∞
Date: October 30, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 17
For the reverse inequality, we need the assumption that f ∈ Lr for
some (ﬁnite) r. For r < p < ∞, we have
1−r/p
�f �p ≤ �f �r/p
.
r �f �∞
Since �f �r < ∞
,
lim sup �f �p ≤ �f �∞ .
p→∞
�
The inequality used in the proof can be written as
�
�
�f �p p
µ({x ∈ X | |f (x)| ≥ t}) ≤
,
t
and is known as Chebyshev’s inequality.
Finite measure spaces. If the measure of the space X is ﬁnite, then
there are inclusion relations between Lp spaces. To exclude trivialities,
we will assume throughout that 0 < µ(X) < ∞.
Theorem 0.2. If q ≤ p < q < ∞, then Lq ⊂ Lp .
Proof. Applying Hölder’s inequality to |f |p and 1,
�
�
p
|f | dµ = |f |p · 1 dµ
��
≤
pq/p
|f |
��
=
�p/q ��
�1−p/q
dµ
dµ
�p/q
|f | dµ
µ(X)1−p/q .
q
�
In particular, if µ(X) = 1, then
�f �1 ≤ �f �p ≤ �f �q ≤ �f �∞ .
Counting measure and lp spaces. Let X be any set, M = P(X),
and µ be the counting measure. Recall that µ(A) is the number of
points in A if A is ﬁnite and equals ∞ otherwise. Integration is simply
�
�
f dµ =
f (x)
X
x∈X
p
for any non­negative function f , and L is denoted by lp .
Theorem 0.3. If 1 ≤ p < q ≤ ∞, then lp ⊂ lq , and
�f �∞ ≤ �f �q ≤ �f �p ≤ �f �1 .
MEASURE AND INTEGRATION: LECTURE 17
3
Proof. If q = ∞, then observe that for any x0 ∈ X,
�
�1/p
�
p
|f (x0 )| ≤
|f (x)|
.
x∈X
Now let q < ∞. Then we NTS
�
�1/q �
�1/p
�
�
q
p
|f (x)|
≤
|f (x)|
.
x∈X
x∈X
Now multiply both
is equal to 1.
� sides pby a constant so that�the RHS
q
Thus, assuming
|f (x)| = 1, we NTS that
|f (x)| ≤ 1. But this
is immediate, since |f (x)| ≤ 1 for all x implies that |f (x)|q ≤ |f (x)|p
because q > p.
�
Thus, in a certain sense, the counting measure and a ﬁnite measure
act in reverse ways for Lp spaces.
Local LP spaces. Let G be an open set in Rn . The local Lp space on
G consists of all L­measurable functions f deﬁned a.e. on G such that
for every compact set K ⊂ G, the characteristic function f χK has a
ﬁnite Lp norm; that is,
�
|f (x)|p dx < ∞ if 1 ≤ p < ∞;
K
f is essentially bounded on K
if p = ∞.
This set is denoted Lploc (G). From our result on ﬁnite measure spaces,
we have at once for 1 ≤ p < q ≤ ∞,
p
q
1
L∞
loc (G) ⊂ Lloc (G) ⊂ Lloc (G) ⊂ Lloc (G).
Convexity properties of Lp norm. Let (X, M, µ) be a measure
space.
Theorem 0.4. Let 1 ≤ p < r < q < ∞ and suppose f ∈ Lp ∩ Lq .
Then f ∈ Lr and
log �f �r ≤
1
r
1
p
−
−
1
q
1
q
log �f �p +
1
p
1
p
−
−
1
r
1
q
log �f �q .
Proof. Since 1/q < 1/r < 1/p, there exists a unique θ such that
1
θ 1−θ
= +
.
r
p
q
4
MEASURE AND INTEGRATION: LECTURE 17
The number θ satisﬁes 0 < θ < 1 and equals
θ=
1
r
1
p
−
−
1
q
1
q
,
1−θ =
1
p
1
p
−
−
1
r
1
q
.
We NTS that log �f �r ≤ θ log �f �p + (1 − θ) log �f �q . Note that
rθ r(1 − θ)
+
,
p
q
and so p/rθ and q/r(1 − θ) are conjugate exponents. Thus, by Hölder’s
inequality,
�
�
�f �r = �
f θ f 1−θ �
r
�
�1/r
=
�
f rθ f r(1−θ) �
1
�1/r
�� �
�
�
≤ �
f rθ �
p/rθ �
f r(1−θ) �
q/r(1−θ)
�1/r
�
r(1−θ)
= �f �rθ
�f
�
p
q
1=
= �f �θp �f �1−θ
.
q
�
The theorem states that if f is an M­measurable non­zero function
on X, then the set of indices p such that f ∈ Lp is an interval I ⊂ [1, ∞],
and log �f �p is a convex function of 1/p on I.
MEASURE AND INTEGRATION: LECTURE 18
Fubini’s theorem
Notation. Let � and m be positive integers, and n = � + m. Write Rn
as the Cartesian product Rn = R� + Rm . We will write points in Rn as
z ∈ Rn ;
x ∈ R� ;
y ∈ Rm ;
z = (x, y).
If f is a function on Rn and y ∈ Rm is ﬁxed, then fy is the function on
R� deﬁned by
fy (x) = f (x, y).
The function fy is called the section of f determined by y. In particular,
if A ⊂ Rn and f = χA , then
�
1 if (x, y) ∈ A;
fy (x) =
0 if (x, y) �∈ A.
In this case, fy is the characteristic function of a subset of R� , and a
point x ∈ R� is in this set if and only if (x, y) ∈ A. This set will be
denoted by
Ay = {x ∈ R� | (x, y) ∈ A},
and is called the section of A determined by y.
Now let f be any function on Rn . For a ﬁxed y ∈ Rm , it may be
that the function fy on R� is integrable. In this case, let
�
F (y) =
fy (x) dx.
R�
Of course, fy must be L­measurable, but there are two ways F (y) could
exist: (1) fy ≥ 0, in which case 0 ≤ F (y) ≤ ∞, and (2) fy ∈ L1 (R� ),
in which case −∞ < F (y) < ∞.
We eventually want to prove the equation
�
�
F (y) dy =
f (z) dz.
Rm
Rn
To show this, we assume f is L­measurable and integrable, and prove
that F (y) exists for a.e. y ∈ Rm and that F is L­measurable and
integrable on Rm .
Date: November 4, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 18
However, it cannot be expected that fy is an L­measurable function
for all y ∈ Rm . Indeed, let E ⊂ R� be a non­measurable set, ﬁx
� y0 but Ay0 = E.
y0 ∈ Rm , and let A = E × {y0 }. Then Ay = ∅ if y =
The set A is measurable with λ(A) = 0. But Ay0 is not measurable.
Fubini I: Non­negative functions.
Theorem 0.1. Assume that f : Rn → [0, ∞] is L­measurable. Then
for a.e. y ∈ Rm , the function fy : R� → [0, ∞] is L­measurable, and so
�
F (y) =
fy (x) dx
R�
exists. Moreover, F is L­measurable on Rm , and
�
�
F (y) dy =
f (x) dz.
Rm
Rn
The second equation will be abbreviated
�
� ��
�
f (x, y) dx dy =
f (x, y) dx dy,
Rm
R�
Rn
and the LHS of this equation will often be written
� �
�
�
f (x, y) dx dy or
dy
f (x, y) dx.
Rm
R�
Rm
R�
Proof. The proof is long, and is broken into 10 steps.
(1) Let J be a special rectangle. Then J = J1 × J2 , with J1 and J2
special rectangles in R� and Rm . Then for any y ∈ Rm ,
�
J1 if y ∈ J2 ;
Jy =
∅
if y �∈ J2 .
Thus, λ(Jy ) = λ(J1 )χJ2 (y), and so
�
λ(Jy ) dy = λ(J1 )λ(J2 )
Rm
= λ(J ).
(2) Let G ⊂ Rn be open, and write G = ∪∞
k=1 Jk , with each Jk a
disjoint rectangle. Thus,
Gy =
∞
k=1
Jk,y
MEASURE AND INTEGRATION: LECTURE 18
3
�
is a disjoint union, and so λ(Gy ) = λ(Jk,y ). Thus,
�
∞ �
�
λ(Gy ) dy =
λ(Jk,y ) dy
Rm
=
k=1
∞
�
Rm
λ(Jk )
k=1
= λ(G).
(3) Let K ⊂ Rn be compact, and choose G ⊃ K open and bounded.
Apply (2) to G \ K:
�
λ(Gy \ Ky ) dy = λ(G \ K);
Rm
�
�
λ(Gy ) dy −
λ(Ky ) dy = λ(G) − λ(K).
Rm
Rm
Thus, applying (2) to G gives
�
λ(Ky ) dy = λ(K).
Rm
(4) Let K1 ⊂ K2 ⊂ · · · be compact. Let B = ∪k Kj . Then for all
y ∈ Rm ,
∞
By =
Kj,y .
j=1
So By is measurable, λ(By ) = limj → ∞λ(Kj,y ) is increasing,
so by monotone convergence,
�
�
λ(By ) dy = lim
λ(Kj,y ) dy
j→∞
Rm
Rm
= lim λ(Kj )
j→∞
by (3)
= λ(B).
(5) Let G1 ⊃ G2 ⊃ · · · be open and bounded. Let C = ∩j Gj and
let K ⊃ G1 . Applying (4),
∞
K \C =
(K \ Gj ),
j=1
and so
Since
�
Rm
λ(Ky \ Cy ) dy = λ(K \ C).
�
Rm
λ(Ky ) dy = λ(K),
4
MEASURE AND INTEGRATION: LECTURE 18
the result follows for C.
(6) This step is the most important. Let A be bounded and mea­
surable. By the approximation theorem, there exist compact
sets Kj and bounded open sets Gj such that
K1 ⊂ K 2 ⊂ · · · ⊂ A ⊂ · · · ⊂ G 2 ⊂ G 1
and
lim λ(Kj ) = λ(A) = lim λ(Gj ).
j→∞
j→∞
Let
∞
B=
Kj
and C =
j=1
∞
�
.
j=1
Then B ⊂ A ⊂ C and λ(B) = λ(A) = λ(C). Thus, by (4) and
(5),
�
(λ(Cy ) − λ(By )) dy = 0,
Rm
and so λ(Cy ) − λ(By ) = 0 for a.e. y ∈ Rm . This means that
Cy \ By has measure zero in R� for a.e. y ∈ Rm , and for these y,
By ⊂ Ay ⊂ Cy ⇒ Ay = By ∪ N , where N is a null set. Hence,
Ay is measurable for a.e. y, λ(Ay ) is a measurable function of
y, and
�
�
λ(Ay ) dy =
λ(By ) dy
Rm
Rm
= λ(B) = λ(A).
(7) Observe that if the theorem is valid for each function 0 ≤ f1 ≤
f2 ≤ · · · , then it is valid for f = lim fj . This is due to monotone
convergence, (2) and (4). Since fj,y is measurable for a.e. y, fy
is L­measurable for a.e. y, and thus for a.e. y ∈ Rm ,
�
F (y) =
fy (x) dx
R�
�
= lim
fj,y (x) dx
j→∞
R�
= lim Fj (y).
j→∞
MEASURE AND INTEGRATION: LECTURE 18
5
Since this is an increasing limit and Fj is measurable, so is F ,
and by monotone convergence,
�
�
F (y) dy = lim
Fj (y) dy
j→∞ Rm
Rm
�
= lim
fj (z) dz
j→∞ Rn
�
=
f (z) dz.
Rn
(8) Let fj be the characteristic function of the bounded set A ∩
B(0, j). Then the theorem is valid for the characteristic func­
tion of any measurable set by (6) the observation in (7).
(9) Since non­negative measurable simple functions are (ﬁnite) lin­
ear combinations of functions in (8), the theorem follows for
them.
(10) The theorem follows from the theorem that states that there
exists a sequence of simple measurable functions converging to
any non­negative measurable function.
�
Fubini II: Integrable functions.
Theorem 0.2. Assume that f ∈ L1 (Rn ). Then for a.e. y ∈ Rm , the
function fy ∈ L1 (R� ), and
�
F (y) =
fy (x) dx
R�
exists. Moreover, F ∈ L1 (Rm ), and
�
�
F (y) dy =
Rm
Rn
f (z) dz.
Proof. Write f = f + − f − and apply Fubini I. Deﬁne
�
�
−
fy dx, H(y) =
fy+ dx,
G(y) =
R�
so that
�
Rm
R�
�
G dy =
Rn
f
−
�
dz,
Rm
�
H dy =
Rn
f + dz.
Because the integrals are ﬁnite, G(y) < ∞ and H(y) < ∞ a.e. and
thus fy ∈ L1 (R� ). Also, F (y) = H(y) − G(y) a.e., and so F ∈ L1 (Rm )
6
MEASURE AND INTEGRATION: LECTURE 18
and
�
Rm
�
F dy =
=
=
�
�R
m
�R
n
Rn
H dy −
f + dz −
�R
m
Rn
G dy
f − dz
f dz.
�
Example of Fubini’s theorem. Let us calculate the integral
�
y sin xe−xy dx dy,
E
where E = (0, ∞) × (0, 1). Since the integrand is a a continuous func­
tion, it is L­measurable. We have by integration by parts
� ∞
F (y) =
y sin xe−xy dx
0
=
Thus,
�
1
y2
y
.
+1
1
log 2.
2
0
Now, since |f (x, y)| ≤ ye−xy , we may apply Fubini I to see that
�
�
|f (x, y)| dx dy ≤
ye−xy dx dy
E
E
� 1
� ∞
=
dy
ye−xy dx
0
0
� 1
=
dy
F (y) dy =
0
= 1.
Doing integration with respect to y ﬁrst yields
�
�
� 1
sin x 1 − e−x
−x
−xy
.
−e
y sin xe
dy =
x
x
0
Thus, Fubini’s theorem shows that
�
�
� ∞
1
sin x 1 − e−x
−x
dx = log 2.
−e
x
x
2
0
MEASURE AND INTEGRATION: LECTURE 19
Product spaces in Rn .
Proposition 0.1. Let Rn = R� × Rm . Let X ⊂ R� be L� ­measurable
and Y ⊂ Rm be Lm ­measurable. Then X × Y ⊂ Rn is Ln ­measurable,
and λ(X × Y ) = λ(X)λ(Y ).
Proof. If X × Y ∈ Ln , then by Fubini I,
�
�
λ(X × Y ) =
χ(X×Y ) dz =
χX χY dz
Rn
Rn
�
� ��
=
χX χY dx dy = λ(X)λ(Y ).
Rm
R�
Just NTS X × Y ∈ Ln .
We may assume X and Y have ﬁnite measure. Let X = ∪∞
k=1 Xk and
∞
Y = ∪k=1 Yk , where Xk = X ∩ B(0, k) and Yk = Y ∩ B(0, k). Then
X ×Y =
∞
Xk × Yk .
j,k
n
n
So if Xk × Yk ∈ L , since L is a σ­algebra, then X × Y ∈ Ln .
Now, given � > 0, there exists K1 ⊂ X ⊂ G1 and K2 ⊂ Y ⊂ G2 ,
with K1 ⊂ R� and K2 ⊂ Rm compact, G1 ⊂ R� and G2 ⊂ Rm open,
such that λ� (G1 \K1 ) < � and λm (G2 \K2 ) < �. We have K1 ×K2 ⊂ Rn
is compact, G1 × G2 ⊂ Rn is open, and K1 × K2 ⊂ X × Y ⊂ G1 × G2 .
Now
G1 × G2 \ K1 × K2 = ((G1 \ K1 ) × G2 ) ∪ (K1 × (G2 \ K2 ))
⊂ ((G1 \ K1 ) × G2 ) ∪ (G1 × (G2 \ K2 )) .
Thus,
λ(G1 × G2 \ K1 × K2 ) = λ(G1 \ K1 )λ(G2 ) + λ(G1 )λ(G2 \ K2 )
≤ �λ(G2 ) + �λ(G1 )
< �(λ(K2 ) + �) + �(λ(K1 ) + �)
≤ �(λ(X) + λ(Y ) + 2�).
Date: November 6, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 19
Hence λ(G1 × G2 \ K1 × K2 ) can be made arbitrarily small. By the
�
approximation theorem, X × Y is Ln ­measurable.
General product spaces. Let (X, MX , µX ) and (Y, MY , µY ) be mea­
sure spaces. What is a measure on X × Y ? Deﬁne MX × MY to be
the smallest σ­algebra containing measurable rectangles (i.e., A × B
with A ∈ MX and B ∈ MY ).
Proposition 0.2. If E ∈ MX × MY , then Ey (the x­section of E at
y) is in MX for all y ∈ Y .
Proof. Let Ω be the class of all E ∈ MX × MY such that Ey ∈ MX
for every y ∈ Y . If E = A × B, then clearly E ∈ Ω. Then Ω is a
σ­algebra: (a) X × Y ∈ Ω, (b) E ∈ Ω, then (E c )y = (Ey )c ∈ MX since
∞
MX is a σ­algebra. (c) If Ei ∈ Ω, then (∪∞
i=1 Ei )y = ∪i=1 (Ei )y ∈ MX
since MX is a σ­algebra.
�
For Rn = R� × Rm , it is not true that Ln is the product measure
(but it is the completion of the product measure). How do we deﬁne
µX ×Y ?
Proposition 0.3. If E ∈ MX × MY , then Ey ∈ MX for all y and
λ(Ey ) is a measurable function on Y .
�
Deﬁne λX×Y (E) = Y λ(Ey ) dµY . If X and Y are σ­ﬁnite
� (countable
unions of sets with ﬁnite measure), then this also equals X λ(EX ) dµX .
Fubini’s theorem. Let (X, MX , µX ) and (Y, MY , µY ) be σ­ﬁnite mea­
sure spaces and f = MX × MY measurable. Then, for each y ∈ Y , fy
is MX ­measurable, and for each x ∈ X, fx is MY ­measurable.
(a) Let 0 ≤ f ≤ ∞,
�
�
ϕ(x) =
fx dµY , ψ(y) =
fy dµX .
Y
X
Then ϕ is MX ­measurable and ψ is MY ­measurable, and
�
�
�
ϕ dµX =
f d(µX × µY ) =
ψ dµY .
X
X×Y
Y
(b) Let f : X × Y → C. If f ∈ L1 (µX × µY ), then fX ∈ L1 (µY )
for a.e. x ∈ X and fY ∈ L1 (µX ) for a.e. y ∈ Y , and the above
holds (ϕ ∈ L1 (µX ) and ψ ∈ L1 (µY )).
If µX and µY are complete and use µX × µY (the completion of
µX × µY , then the only change is fY is MX ­measurable for a.e. y and
fX is MY ­measurable for a.e. x.
Proposition 0.4. Let f : X × Y → C is MX × MY ­measurable. Then
MEASURE AND INTEGRATION: LECTURE 19
3
(a) for every x ∈ X, fx : Y → C is MY ­measurable,
(b) for every y ∈ Y , fy : X → C is MX ­measurable.
Proof. If V is open, let Q = f −1 (V ), Q ∈ MX × MY . We have
�
Qx = {y | fx (y) ∈ V } = fx−1 (V ) ∈ MY from earlier.
Theorem 0.5. Let Rn = R� × Rm . Then Ln is the completion of
L� × Lm .
MEASURE AND INTEGRATION: LECTURE 20
Convolutions
Deﬁnition. If f and g are measurable functions on Rn , then the con­
volution of f and g, denoted f ∗ g, is deﬁned formally as
�
(f ∗ g)(x) =
f (y)g(x − y) dy.
Rn
The operation is commutative and associative:
(f ∗ g)(x) = (g ∗ f )(x) and (f ∗ g) ∗ h = f ∗ (g ∗ h).
Inequalities. Let f be a Lebesgue measurable function on Rn . Then
the function f (x) considered as a function of (x, x) in R2n is Lebesgue
measurable since Ln × Ln ⊂ L2n . The linear transformation given by
(x, y) �→ (x−y, y) is invertible, and so f (x−y) is a Lebesgue measurable
function of (x, y) ∈ R2n . Thus, we see that f (y)g(x − y) is measurable
on R2n .
The next theorem asserts that if f and g are in L1 (Rn ), then f ∗ g
exists a.e. and f ∗ g ∈ L1 (Rn ). Since the product of two integrable
functions need not be integrable, it is not obvious that f ∗ g exists a.e.
Theorem 0.1. Assume f, g ∈ L1 (Rn ). Then for a.e. x ∈ Rn , the
convolution (f ∗ g)(x) exists, f ∗ g ∈ L1 (Rn ), and
�f ∗ g�1 ≤ �f �1 �g�1 .
Proof. Assume that f and g are non­negative. Then f (y)g(x − y) is a
non­negative measurable function, and Fubini I implies
�
�
�
�
dx f (y)g(x − y) dy =
dy f (y)g(x − y) dx.
�
The LHS equals (f ∗ g)(x) dx, and the RHS is
�
�
�
�
f (y) dy g(x − y) dx = f (y) dy · g(x) dx.
Thus �f ∗ g�1 = �f �1 �g�1 . When f and g are not necessarily non­
negative, we see that |f | ∗ |g| exists a.e. ⇒ |f (y)g(x − y)| integable
⇒ f (y)g(x − y) integrable ⇒ f ∗ g exists a.e. Since |f ∗ g| ≤ |f | ∗ |g|,
the theorem follows.
�
Date: November 13, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 20
Young’s theorem. Our next theorem generalizes the previous one.
Theorem 0.2. Let p, q, r ∈ [1, ∞] such that
1
1 1
= + − 1.
r
p q
If f ∈ Lp (Rn ) and g ∈ Lq (Rn ), then f ∗g exists a.e. and f ∗g ∈ Lr (Rn ).
Moreover,
�f ∗ g�r ≤ �f �p �g�q .
Proof. Without loss of generality, let �f �p = �g�p = 1. The general
case follows from the non­negative case, so assume f, g ≥ 0. Applying
Hölder’s inequality,
�
�
�
(f ∗ g)(x) =
f (y)p/r g(x − y)q/r f (y)1−p/r g(x − y)1−q/r dy
��
≤
p
q
�1/r ��
(1−p/r)q �
f (y) g(x − y) dy
��
×
(1−q/r)p�
g(x − y)
f (y)
�1/p�
dy
.
We have used the fact that
�
� �
�
1
1
1
1
1
1
+ + = + 1−
+ 1−
= 1.
r q � p�
r
q
p
Since
�
�
�
�
p� �
1 1 �
1
−
q =p
q =p 1−
= p,
1−
p r
q
r
�
�
�
�
�
q� �
1 1 �
1 �
p =q
−
p =q 1−
p = q,
1−
q r
p
r
�
we have
��
(f ∗ g)(x) ≤
q
f (y) g(x − y) dy
i.e.,
r
p
(f ∗ g) (x) ≤
�
�1/r
· 1 · 1,
f (y)p g(x − y)q dy.
�1/q�
dy
MEASURE AND INTEGRATION: LECTURE 20
3
Thus, (f ∗ g)r ≤ f p ∗ g q , and so
�
(f ∗ g) dx ≤ �f p ∗ g q �1
= �f p �1 �g q �1
= �f �pp �g�qq
= 1.
�
The proof ignores the case in which some of the exponents equal ∞.
But, if p = ∞, then r = ∞ and q = 1, and the result follows since
|f ∗ g| ≤ �f �∞ �g�1 . If r = ∞, then q = p� , and the result follows from
H¨older’s inequality. However, more is true when r = ∞.
Theorem 0.3. Let 1 ≤ p ≤ ∞ and f ∈ Lp (Rn ). Then the integral
deﬁning (f ∗ g)(x) exists for all x ∈ Rn , f ∗ g is bounded and uniformly
continuous, and if 1 < p < ∞, then f ∗g ∈ C0 (i.e., lim|x|→∞ (f ∗g)(x) =
0).
Proof. Either p or p� must be ﬁnite. Suppose p� < ∞. The corollary
to Cc dense in Lp implies that for all � > 0 there exists δ > 0 such
that if |y | < δ, then �τy g − g�p� ≤ �, where τ is translation by y. Thus,
|x − x� | ≤ δ, then
�τx g − τx� g�p� = �τx−x� g − g�p� ≤ �.
By Hölder’s inequality,
�
�
|(f ∗ g)(x) − (f ∗ g)(x )| ≤
=
�
|f (y)| |g(x − y) − g(x� − y| dy
|f (−y)| |g(x + y) − g(x� + y)| dy
≤ �f �p �τx g − τx� g�p�
≤ �f �p �.
This proves that f ∗ g is uniformly continuous.
Now let 1 < p < ∞. Since Cc is dense in Lp , there exist sequences
�
fk , gk ∈ Cc (Rn ) such that fk → f in Lp and gk → g in Lp . Thus,
fk ∗ gk ∈ Cc (Rn ). Estimating,
�fk ∗ gk − f ∗ g�∞ ≤ �fk ∗ (gk − g)�∞ + �(fk − f ) ∗ g�∞
≤ �fk �p �gk − g�p� + �fk − f �p �g�p6�
→ 0 as k → ∞.
Thus, fk ∗ gk converges uniformly to f ∗ g, and so f ∗ g → 0 as
|x| → ∞.
�
MEASURE AND INTEGRATION: LECTURE 21
Approximations. Let
�
0
t ≤ 0;
h(t) =
exp(−1/t) t > 0.
Then h ∈ C ∞ (inﬁnitely diﬀerentiable with continuous derivatives).
Deﬁne φ : Rn → R by φ(x1 , . . . , xn ) = h(1 − |x|2 ). If |x|2 > 1, then
1 − |x�|2 < 0 ⇒ φ = 0 on B(0, 1)c . Thus, φ ∈ Cc∞ (Rn ). Redeﬁne φ so
that Rn φ dz = 1.
Now deﬁne φa (x) = a−n φ(x/a). Then φa supported on a ball of
�
Rn
φa (x)dx = 1
by a linear change of variables.
�
Given f , deﬁne fa (x) = f ∗ φa = Rn f (y)φa (x − y) dy. Then fa (x) ∈
C0∞ since
�
∂ (k)
∂ (k)
f
(x)
=
f
(y)
φa (x − y) dy,
a
∂x(k)
∂x(k)
Rn
and if f has compact support, then so does fa .
Suppose f ∈ L1 (Rn ) and deﬁne
�
g(x) =
f (y)φa (x − y) dy = f ∗ φa .
Rn
Note that φa (x − y) is bounded and the integrand is integrable.
Lemma 0.1. The function g(x) is continuous.
Proof. Fix x0 . Then
�
lim g(x) = lim
x→x0
Date: November 18, 2003.
x→x0
Rn
1
f (y)φa (x − y) dy,
2
MEASURE AND INTEGRATION: LECTURE 21
and since f (y)φa (x − y) ≤ C |f (y)| ∈ L1 , we may apply LDCT so that
the RHS above equals
�
=
lim f (y)φa (x − y) dy
Rn x→x0
�
=
f (y)φa (x − y) dy since φa ∈ Cc
Rn
= g(x0 ).
�
Lemma 0.2. The kth partial derivatives of g exist and are continuous
for k = 1, 2, . . .. In other words, g ∈ C ∞ .
Proof. Let ek = (0, . . . , 0, 1, 0, . . . , 0), the vector whose kth coordinate
is equal to 1 and all other coordinates are zero. We have
�
�
f
(y)φ
(x
+
te
−
y)
dy
−
f (y)φa (x − y) dy
g(x + tek ) − g(x)
a
k
n
Rn
= R
t
t
�
�
�
φa (x + tek − y) − φa (x − y)
=
f (y)
dy.
t
Rn
Since
�
�
φa (x + tek − y) − φa (x − y)
∂k
�
�
=
φ
(x
−
y)
a
�
�
t
∂xk
x =x+t� ek , 0≤t� ≤t
is less than some constant C in absolute value, the integrand above is
dominated by C |f | ∈ L1 . Thus,
∂g
g(x + tek ) − g(x)
= lim
∂xk t→0
t
�
�
�
φa (x + tek − y) − φa (x − y)
=
f (y) lim
dy
t→0
t
Rn
�
∂
=
f (y)
φa (x − y) dy.
∂xk
Rn
Thus, the partial derivatives exist, and
∂
φa (x − y) ∈ C 0 ,
∂xk
so by the ﬁrst lemma, ∂g/∂xk is also continuous. By induction, we can
�
conclude that g(x) ∈ C ∞ .
Lemma 0.3. If f ∈ Cc (Rn ), then g ∈ Cc (Rn ).
MEASURE AND INTEGRATION: LECTURE 21
3
Proof. There exists R > 0 such that f = 0 on B(0, R)c . Choose x so
� 0. If
that g(x) �= 0. Then there exists y such that f (y)φa (x − y) =
f (y) =
� 0, then y ∈ B(0, R). If φa (x − y) =
� 0, then x − y ∈ B(0, a).
Thus,
|x| = |x − y + y | ≤ |x − y | + |y | ≤ a + R,
and so g(x) = 0 if |x| ≤ R + a. In other words, g ∈ Cc∞ (Rn ).
�
Theorem 0.4. Cc∞ is dense in Lp .
Proof for L1 . We proved previously that Cc is dense in L1 ; we just need
to prove that Cc∞ is dense in Cc . Given f ∈ Cc , there exists r > 0 such
that f = 0 on B(0, r)c . Given � > 0, since f ∈ Cc , f is uniformly
continuous means that there exists a > 0 such that |x − y | ≤ a ⇒
�
|f (x) − f (y)| ≤
,
λ(B(0, r + 1))
and we may make 0 < a ≤ 1.
Consider φa :
�
�
φa dx = 1 and
φa (x − y)dy = 1.
Thus,
��
�
�
�
|f ∗ φa (x) − f (x)| = ��
(f (y) − f (x))φa (x − y) dy ��
�
≤ |f (y) − f (x)| φa (x − y) dy
�
=
|
f (y) − f (x)| φa (x − y) dy
|x−y|≤a
�
�
φa (x − y) dy
≤
λ(b(0, r + 1)) |x−y|≤a
�
=
.
λ(B(0, r + 1))
So we have that
�
�f ∗ φa − f �1 =
=
≤
Rn
�
|f ∗ φa (x) − f (x)| dx
B(0,r+1)
|f ∗ φa (x) − f (x)| dx
�
λ(B(0, r + 1)) = �.
λ(B(0, r + 1)
�
In fact, more is true. We ﬁrst need a lemma.
4
MEASURE AND INTEGRATION: LECTURE 21
Lemma 0.5. If f ∈ L1 (Rn ), then
�
lim
|f (x + y) − f (x)| dx = 0.
y→0
Rn
Theorem 0.6. Let f ∈ Lp (Rn ) with 1 ≤ p < ∞. Then
lim �f ∗ φa − f �p = 0.
a→0
Proof for L1 . We have
�
�
� ��
(f ∗ φa (x) − f (x)) =
(f (x − y) − f (x))φa (y) dy dx
�
� ��
=
(f (x − y) − f (x))φa (y) dx dy
�
�
= φa (y) (f (x − y) − f (x)) dx dy
�
�
≤
φa (y) · � +
2 �f �1 φa (y)
B(0,r)
B(0,r)
�
≤ � + 2 �f �1
φa (y)
B(0,r)
→0
for a suﬃciently small.
�
MEASURE AND INTEGRATION: LECTURE 22
Fundamental theorem
of calculus. Let f : [a, b] → R be continuous
�x
and let F (x) = a f (t) dt (a ≤ x ≤ b). Then F is diﬀerentiable and
F � (x) = f (x). But what if f ∈ L1 only? Claim: F is continuous if
f ∈ L1 . Proof:
�
F (x) =
f (t)χ[a,x] dt
[a,b]
�
lim F (x) =
lim f (t)χ[a,x] dt by LDCT
x→x0
[a,b] x→x0
�
=
f (t)χ[a,x0 ] dt = F (x0 ).
[a,b]
Lebesgue’s theorem. If f ∈ L1 , then F is diﬀerentiable a.e. and
F � (x) = f (x) a.e.
Other half of FTOC. If given f diﬀerentiable a.e. (so f � exists), and
let f ∈ L1 . Then is it true that
� x
f (x) − f (a) =
f � (t) dt?
a
This is not necessarily true. It is true is f is absolutely continuous on
[a, b] or if f is everywhere diﬀerentiable and f � ∈ L1 .
A function f : [a, b] → R is absolutely continuous if for all � > 0 there
exists δ > 0 such that
n
�
|f (βi ) − f (αi )| < �
i=1
for any n and collection
of disjoint segments (α1 , β1 ), . . . , (αn , βn ) in
�
[a, b] such that ni=1 (βi = αi ) < δ. Absolute continuity is stronger
than continuity: continuity is when n = 1.
Theorem 0.1. If f is absolutely continuous, then f is diﬀerentiable
a.e. and f � ∈ L1 . In particular, the second half of the FTOC holds.
Date: November 20, 2003.
1
2
MEASURE AND INTEGRATION: LECTURE 22
Vitali covering theorem.
Theorem 0.2. Let E ⊂ Rn be a bounded set, and suppose we have a
collection of balls F containing E such that every point of E is at the
center of some ball (there may be several balls at each point). Then
there exist balls B1 , B2 , . . . ∈ F so that
(1) B1 , B
2 , . . . are disjoint, and
(2) E ⊂ α≥1 3Bα .
Proof. Assume an upper bound to the radius of the balls (otherwise,
since E is bounded, it is covered by a single ball.) Use the following
selection procedure. Let B1 , . . . , Bk−1 be selected, and deﬁne
dk = sup{rad B | B ∈ F, B ∩
Bj = ∅},
j<k
the largest radius of balls disjoint from the union of previously selected
balls. First step is just the choosing the ball with largest radius. If
no disjoint balls remain, then stop. Otherwise, choose Bk ∈ F disjoint
from B1 , . . . , Bk−1 such that rad Bk > dk /2. This gives a ﬁnite or
countably inﬁnite collection of disjoint balls, so (1) is satisﬁed. To
prove (2), choose any x ∈ E, and we NTS that x ∈ 3Bα for some α.
From the assumption, x ∈ B (at center) for some B. Let ρ = rad B.
Claim: B must hit some Bk in our choice. If not, the selection process
never terminates, so that dk > ρ for all k. But this implies that the
boundedness.
Choose smallest k so that B ∩ Bk =
� ∅, so B ∩ Bj = ∅ for all j < k.
Then ρ ≤ dk < 2 rad Bk . Choose y ∈ B ∩ Bk and let z be the center of
Bα . Then
|x − z | ≤ |x − y | + |y − z |
and thus x ∈ 3Bk .
�
The maximal function. Given f ∈ L1 (Rn ), the maximal function of
f is
�
1
|f (y)| dy.
M f (x) = sup
0<f <∞ λ(B(x, r)) B(x,r)
Let us show that M f is measurable. Let t < M f (x), then there
exists 0 < r < ∞ such that
�
1
t<
|f (y)| dy.
λ(B(x, r)) B(x,r)
MEASURE AND INTEGRATION: LECTURE 22
Choose any r� > r so that
1
t<
λ(B(x, r� ))
3
�
B(x,r)
|f (y)| dy.
Note that if we have x� such that |x − x� | ≤ r� − r, then B(x, r) ⊂
B(x� , r� ). Thus,
�
1
t<
|f (y)| dy
λ(B(x, r� )) B(x� ,r� )
�
1
|f (y)| dy.
≤
λ(B(x� , r� )) B(x� ,r� )
≤ M f (x� ).
We’ve shown that if M f (x) > t, then for x� close to x, M f (x� ) > t.
Hence, M f is lower semicontinuous and thus Borel measurable.
Recall that if g ∈ L1 , then
�
�
�g�1 = |g | ≥
|g | ≥ tµ({x | |
g(x)| ≥ t}).
µ({x | |g(x)|≥t})
Thus, Chebyshev’s inequality,
�g�1
t
holds. If g satisﬁes µ({x | |g(x)| > t}) < c/t, this does not imply that
g ∈ L1 . For example, g(x) = |x|−n on Rn . This condition is known as
weak (1, 1).
µ({x | |g(x)| ≥ t}) ≤
Theorem 0.3. If f ∈ L1 (Rn ), then
λ({x | M f (x) > t}) ≤
3n �f �1
t
for all 0 < t < ∞,
i.e., M f is weak (1, 1).
Remark. M f is not L1 unless f = 0.
Proof of remark. Choose a > 0 and |x| > a. Then
�
1
Mf ≥
|f (y)| dy
B(x, 2 |x|) B(x,2|x|)
�
1
|f (y)| dy
≥
B(0, 2 |x|) B(0,a)
�
C
= n
|f (y)| dy.
x B(0,a)
4
MEASURE AND INTEGRATION: LECTURE 22
If M f (x) is not integrable, then
�
|f (y)| dy = 0 ⇒ f = 0
B(0,a)
since 1/xn is not integrable when x is large.
�
Proof of theorem. Let E = {x | M f (x) > t}. We NTS that
λ(E) ≤ 3n �f �1 /t.
If we show that
λ(E ∩ b(0, k)) ≤
3n �f �1
t
for any k, then by taking limits (i.e., increasing sequence of sets), this
will show that
λ(E) ≤ 3n �f �1 /t.
So, without loss of generality, assume E is bounded.
Given x ∈ E,
�
1
|f (y)| dy > t,
sup
0<r<∞ λ(B(x, r)) B(x,r)
so there exists r such that
1
sup
0<r<∞ λ(B(x, r))
�
B(x,r)
|f (y)| dy > t,
i.e., for any x ∈ E, there exists r such that
�
1
(0.1)
λ(B(x, r)) <
|f (y)| dy.
t B(x,r)
Let F be the collection of all balls centered at E satisfying (0.1).
Then all the hypotheses of the Vitali covering theorem are satisﬁed (E
bounded and every point is at the center of some ball in F). Thus,
there exists a sequence of balls B1 , B2 , . . . ∈ F such that
(1) B1 , B
2 , . . . are disjoint, and
(2) E ⊂ k≥1 3Bk .
MEASURE AND INTEGRATION: LECTURE 22
Thus,
λ(E) ≤
�
k≥1
λ(3Bk ) =
�
3n λ(Bk )
k≥1
�
n1
|f (y)| dy
<
3
t Bk
k≥1
�
n1
=3
|f (y)| dy
t ∪k≥1 Bk
�
n1
≤3
|f (y)| dy.
t Rn
�
5
�
MEASURE AND INTEGRATION: LECTURE 23
Lebesgue’s diﬀerentiation theorem.
Theorem 0.1. Let f ∈ L1 (Rn ). Then for almost every x ∈ Rn ,
�
1
lim
|f (y) − f (x)| dy = 0.
r→0 λ(B(x, r)) B(x,r)
In particular, for a.e. x ∈ Rn ,
�
1
lim
f (y) dy = f (x).
r →0 λ(B(x, r)) B(x,r)
Remark. This looks like the FTOC I almost everywhere: the deriva­
tive of the integral of f = f . Next time, prove this and show it implies
FTOC in the case of R.
Proof of theorem. Obviously,
��
�
�
�
�
�
1
�
�
f
(y)
dy
−
f
(x)
� λ(B(x, r))
�
B(x,r)
�
�
�
�
�
1
�
(f (y) − f (x))��
=�
λ(B(x, r)) B(x,r)
�
1
|f (y) − f (x)| dy.
≤
λ(B(x, r)) B(x,r)
Thus, the particular case of the theorem follows from the ﬁrst state­
ment.
Recall that if f ∈ L1 , we can deﬁne the maximal function M f , and
3n �f �1
λ({x | M f (x) ≥ t}) <
.
t
Also,
∞
�
{x | M f (x) ≥ t} =
{x | M f (x) > t − 1/j},
j=1
so
λ({x | M f (x) ≥ t} ≤
Date: November 25, 2003.
1
3n �f �1
.
t
2
MEASURE AND INTEGRATION: LECTURE 23
Deﬁne
1
f (x) = lim sup
λ(B(x, r))
r →0
∗
�
B(x,r)
|f (y) − f (x)| dy.
We want to show that f ∗ (x) = 0 a.e. The function f ∗ has the following
properties.
(1) f ∗ ≥ 0.
(2) (f + g)∗ ≤ f ∗ + g ∗ .
Proof.
�
B(x,r)
|f (y) + g(y) − f (x) − g(x)| dy
�
=
B(x,r)
�
≤
B(x,r)
|f (y) − f (x) + g(y) − g(x)| dy
�
|f (y) − f (x)| dy +
|g(y) − g(x)| dy.
B(x,r)
�
(3) If g is continuous at x, then g ∗ (x) = 0.
Proof. For any � > 0, there exists δ such that |g(y) − g(x)| ≤ �
for all y ∈ B(x, δ). So, for 0 < r ≤ δ,
�
1
1
�λ(B(x, r)) = �.
|g(y) − g(x)| dy ≤
λ(B(x, r)) B(x,r)
λ(B(x, r))
Thus, g ∗ (x) < � for any �, and hence g ∗ (x) = 0.
�
Note that this is FTOC for continuous functions. The strat­
egy is that we know it is true for continuous functions, so we
will approximate f ∈ L1 (Rn ) by g ∈ C 0 (Rn ).
(4) If g is continuous, then (f − g)∗ = f ∗ .
Proof.
(f − g)∗ ≤ f ∗ + (−g)∗ = f ∗
and
f ∗ ≤ (f − g)∗ + g ∗ = (f − g)∗ .
�
(5) f ∗ ≤ M f + |f |.
MEASURE AND INTEGRATION: LECTURE 23
Proof.
3
�
1
|f (y) − f (x)| dy
λ(B(x, r)) B(x,r)
�
1
≤
(|f (y)| + |f (x)|) dy
λ(B(x, r)) B (x,r)
�
�
�
1
=
|f (y)| dy + |f (x)|
λ(B(x, r)) B(x,r)
≤ M f + |f (x)|
�
(6) Have not proved that f ∗ is measurable, but claim
2(3n + 1)
�f �1
for all 0 < t < ∞.
t
Proof. If f ∗ (x) > t at x, then from (5),
λ∗ ({x | f ∗ (x) > t}) ≤
t < M f (x) + |f | (x),
and so either M f (x) > t/2 or |f (x)| > t/2. So,
{f ∗ > t} ⊂ {M f > t/2} ∪ {|f | > t/2}.
Thus,
∗
λ ({x | f ∗ (x) > t}) ≤ λ({x | M f (x) > t/2}) + λ({x | |f (x)| > t/2})
≤
3n �f �1 �f �1
+
.
t/2
t/2
The last step used the theorem from last time and Chebyshev’s
�
inequality for L1 functions.
To ﬁnish the proof, given � > 0, from the approximation theorem
(Cc (Rn ) dense in L1 ), there exists g ∈ Cc (Rn ) with �f − g�1 ≤ � (f ∗ =
(f − g)∗ .) Thus,
λ∗ ({x | f ∗ (x) > t}) = λ∗ ({x |(f − g)∗ (x) > t})
2(3n + 1)
�f − g�1
t
2(3n + 1)
≤
�.
t
Since � is arbitrary, λ∗ ({x | f ∗ (x) > t}) = 0. In particular, λ∗ ({x | f ∗ (x) >
∗
1/k}) = 0 for all k, and {x | f ∗ (x) > 0} = ∪∞
k=1 {x | f (x) > 1/k}.
Since countable union ⇒ λ({x | f ∗ (x) > 0}) = 0. Since f ∗ ≥ 0, f ∗ = 0
almost everywhere.
�
≤
4
MEASURE AND INTEGRATION: LECTURE 23
Lebesgue set. Let f ∈ L1 (Rn ). Then x ∈ Rn is in the Lebesgue set
of f if there exists a number A such that
�
1
|f (y) − A| dy = 0.
lim
r→0 λ(B(x, r)) B(x,r)
From before, we see that A is unique (for each x).
Note that f does not have to be deﬁned at x in order for x to be
in the Lebesgue set of f . If f = g a.e., then the Lebesgue set of f
coincides with the Lebesgue set of g. If we think of functions in L1 (Rn )
as equivalence classes, then the Lebesgue set of f is well deﬁned.
Lebesgue’s theorem. Almost every x ∈ Rn is in the Lebesgue set of
f , and if f : Rn → R is the representation of equivalence classes [f ],
then A = f (x). For emphasis: If [f ] ∈ L1loc is an equivalence class, then
for x in the Lebesgue set, f (x) is well deﬁned (deﬁned by the above
limit).
For example, let
�
sin(1/x) x �= 0;
g(x) =
0
x = 0.
Then 0 is not in the Lebesgue set of g. Recall: continuous, then in
Lebesgue set.
Regular convergence. A sequence of measurable functions E1 , E2 , . . .
converges regularly to x if there exists c > 0 and r1 , r2 , . . . such that
Ek ⊂ B(x, rk ), limk→∞ rk = 0, and λ(B(x, rk )) ≤ cλ(Ek ) for all k.
Theorem 0.2. Let f ∈ L1 (Rn ), x in the Lebesgue set of f , and
E1 , E2 , . . . converge regularly to x. Then
�
1
f (y) dy.
f (x) = lim
k→∞ λ(Ek E
k
The point here is that we do not have to use balls.
Proof.
��
�
�
1
�
�
λ(Ek )
Ek
�
�
f (y) dy − f (x)
��
≤
�
�
1
|f (y) − f (x)| dy
λ(Ek ) Ek
�
c
|f (y) − f (x)| dy
≤
λ(B(x, rk )) B(x,rk )
→0
as k → ∞ since x is in the Lebesgue set of f .
�
MEASURE AND INTEGRATION: LECTURE 23
5
FTOC II.
Theorem 0.3. Let f ∈ L1 (R) and let a ∈ R. Deﬁne
��
� x
f (y)χ[a,x] (y) dy
R�
F (x) =
f (y) dy =
− R f (y)χ[x,a] (y) dy
a
x ≥ a;
x < a.
Then F is diﬀerentiable a.e. and F � = f a.e.
Proof. Almost every x ∈ R is in the Lebesgue set of f . We show that
F � (x) = f (x) for x in the Lebesgue set of f . By the previous theorem,
�
1
lim
f (y) dy = f (x)
k→∞ λ(Ek ) E
k
for any regular sequence converging to x. Let rk > 0 such that lim rk =
0 and Ek = (x, x + rk ). Then Ek regular and
�
1 x+rk
lim
f (y) dy = f (x),
k→∞ rk x
i.e.,
F (x + rk ) − F (x)
lim
= f (x).
k→∞
rk
Since rk arbitrary,
F (x + h) − F (x)
= f (x),
lim+
h→0
h
and F is right diﬀerentiable. Repeat the argument with Ek = (x−rk , x)
and
F (x + h) − F (x)
lim−
= f (x),
h→0
h
so F is left and right diﬀerentiable and both one­sided derivatives equal
f (x). Thus, F � (x) = f (x) for any x in the Lebesgue set, which is almost
everywhere by the Lebesgue theorem.
�
MEASURE AND INTEGRATION: LECTURE 24
Inequalities
Generalized Minkowski inequality. Let Rn = R� × Rm and z =
(x, y) ∈ Rn . If Rn → C is measurable, then
�
|f (x, y)|p dx : Rm → R = �fy �Lp (R� ) : Rm → R
R�
n
is R ­measurable for 1 ≤ p < ∞.
Assume that
�
�fy �Lp (R� ) dy < ∞.
Rm
�
Then for a.e. x ∈ R , fx (y) : Rm → C is in L1 (Rm ). Let
�
F (x) =
fx (y) dy.
Rm
�
�
Then F (x) : R → C is R ­measurable and we have
�
�F �Lp (R� ) ≤
�fy �Lp (R� ) dy.
Rm
Note this is
�p
�1/p �
��
�1/p
�� � �
�
�
p
�
�
≤
|f (x, y)| dx
dy.
�
m f (x, y) dy
� dx
�
m
�
R
R
R
R
We could replace by X, Y σ­ﬁnite measure space and Y = {p1 , . . . , pn },
dy the counting measure and get old Minkowski’s.
Proof. We have
�
|F (x)| ≤
Rm
|fx (y)| dy,
so without loss of generality, assume f
theorem applies; so now let p > 1.
Deﬁne g : R� × Rm → R≥0 by
⎧
−1/p�
⎪
⎨f (x, y) �fy �Lp (R� )
g(x, y) = 0
⎪
⎩
∞
Then, for each y,
Date: December 9, 2003.
1
≥ 0. If p = 1, then Fubini’s
if 0 < �fy �p < ∞;
if �fy �p = 0;
if �fy �p = ∞.
2
MEASURE AND INTEGRATION: LECTURE 24
(1) f (x, y) ≤ g(x, y) �fy �p1/p
(2)
�
for a.e. x, and
�gy �Lp (R� ) = �fy �1/p
p .
Then
�
F (x) =
≤
Rm
�
f (x, y) dy
Rm
�
g(x, y) �fy �p1/p dy
��
≤ �gx �Lp (Rm )
�1/p�
�fy �p dy
�
= �gx �Lp (Rm ) · C 1/p ,
�
where C = Rm �fy �p dy.
We now use Fubini’s theorem:
�F (x)�pLp (R� )
≤C
p/p�
� �
�gx �pLp (Rm )
�
dx
�
�
��
p−1
p
=c
g(x, y) dy dx
R�
Rm
�
� ��
p−1
p
=c
g(x, y) dx dy
Rm
R�
�
p−1
=c
�gy �pLp (R� ) dy
m
�R
= cp−1
�fy �Lp (R� ) dy
Rm
�
p−1
p
�fy �Lp (R� ) dy.
=c c=c =
R�
Rm
Thus,
�
�F (x)�Lp (R� ) ≤
and so
��
�
�
�
�
�
f
(x,
y)
dy
�
m
�
p
R
L (R� )
Rm
�fy �Lp (R� ) dy
�
≤
Rm
�fy �Lp (R� ) dy.
�
MEASURE AND INTEGRATION: LECTURE 24
3
Application. Let f ∈ L1 (Rn and g ∈ Lp (Rn ). Then f ∗ g ∈ Lp (Rn )
since
�p
�� ��
�1/p
�
�
�
�
�
n f (y)g(x − y) dy
�
dx
Rn
R
�1/p
� ��
p
≤
|f (y)g(x − y)| dx
dy
Rn
Rn
�
=
Rn
��
|f (y)|
=
Rn
|g(x − y)| dx
Rn
�
�1/p
p
dy
|
f (y)| �g�p dy
= �f �1 �g�p .
Distribution functions. Suppose f : X → [0, ∞] and let µ{f > t} =
µ(x | f (x) > t}.
Theorem 0.1.
�
and
�
X
�
f dµ =
�
p
X
f dµ = p
∞
0
∞
0
µ{f > t} dt
µ{f > t}tp−1 dt
More generally, if ϕ is diﬀerentiable, then
�
� ∞
ϕ ◦ f dµ =
µ{f > t}ϕ� (t) dt.
X
Proof. We have
0
�
Rn
|f | dx =
Rn
�
�
=
n
�R �
=
Then
�
Rn
��
�
|f | dx =
=
Rn
R
�
p
∞
�
0 ∞
0
�
=p
0
R
0
|f (x)|
�
dt
dx
χ[0,f (x)] (t) dt dx
χ[0,f (x)] (t) dx dt
µ{|f |p > t} dt
µ{|t| > t1/p } dt
∞
µ{|f | > s}sp−1 ds,
4
MEASURE AND INTEGRATION: LECTURE 24
letting s = t1/p , so sp = t and dt = psp−1 ds.
�
Marcinkiewicz interpolation. Recall the maximal function
�
1
|f (y)| dy.
M f (x) = sup
0<r<∞ λ(B(x, r)) B(x,r)
Note if f ∈ L∞ , then �M f �∞ ≤ �f �∞ . Thus, M maps L∞ into itself:
M : L ∞ → L∞ .
On the other hand, by Hardy­Littlewood, if f ∈ L1 , then
(0.1)
µ{M f > t} ≤
3n
�f �1 ,
t
and M maps L1 to weak L1 .
Using a method called Marcinkiewicz interpolation, we prove the
following.
Theorem 0.2. Let 1 < p < ∞ and f ∈ Lp . Then M f ∈ Lp , and
(0.2)
�M f �p ≤ C(n, p) �f �p ,
where C(n, p) is bounded as p → ∞ and C(n, p) → ∞ as p → 1.
Proof. Observe that M f = M |f |, so assume f ≥ 0. Choose a constant
0 < c < 1 (we will choose the best c later). For t ∈ (0, ∞), write
f = gt + ht , where
�
f (x) f (x) > ct;
gt (x) =
0
f (x) ≤ ct.
So, 0 ≤ ht (x) ≤ ct for every x, and thus ht ∈ L∞ . We have
M f ≤ M gt + M ht ≤ M gt + ct
from (0.2). Thus, M f − ct ≤ M gt , o if M f (x) > t, then (1 − c)t ≤
M gt (x).
Let Et = {f > ct}. Then
λ{M f > t} ≤ λ{M gt > (1 − c)t}
3n
�gt �1
≤
from (0.1)
(1 − c)t
�
3n
=
f dx.
(1 − c)t Et
MEASURE AND INTEGRATION: LECTURE 24
Thus,
�
Rn
p
∞
λ{M f > t}tp−1 dt
0
��
�
� ∞
3n p
p−2
t
f dx dt
≤
1−c 0
Et
�
� �
� f (x)/c
3n p
p−2
=
f (x)
t
dt dx
1 − c Rn
0
�
�p−1
�
1
f (x)
3n p
dx
=
f (x)
1 − c Rn
p−1
c
�
3n p
c1−p
=
·
f (x)p dx
1 − c p − 1 Rn
= C(n, p) �f �pp .
(M f ) dx = p
Thus,
�
5
1/p
3n pc1−p
�M f �p ≤
�f �p .
(1 − c)(p − 1)
�
��
�
→1 as p→∞
�
Choose c = 1/p = (p − 1)/p; this gives the best constant.
�
```
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