MATH 650.THE RADON-NIKODYM THEOREM

This note presents two important theorems in Measure Theory,the

Lebesgue Decomposition and Radon-Nikodym Theorem.They are not

treated in the textbook.

1.Closed subspaces

Let H be a Hilbert space with inner product h¢;¢i and norm k ¢ k.

De¯nition 1.A subspace of H is a subset of H such that,if it contains

x and y,then it also contains ®x+¯y,for every pair of complex numbers

® and ¯.

A closed subspace is a subspace which is also a closed subset of H,

that is,every Cauchy sequence in the subspace converges to a vector

in the subspace.

Exercise 1.Let H = L

2

(X;¹).Show that the collection of simple

functions is a subspace of H,but not closed in general.

Example 1.Let H = L

2

(X;¹),where X = [0;1] and ¹ is Lebesgue

measure.The following are examples of subspaces of H.

(1) the set of all functions f 2 H such that f(x) = f(¡x) for

almost every x 2 X.

(2) the set of all functions f such that

R

f ¢ ¹ = 0.

(3) the set of all functions f which are bounded on a subset of X

of measure ¹(X).

The ¯rst two examples are closed subspaces,but the last one is not.

Example 2.Let H be the set of sequences of complex numbers fz

n

g

such that

P

1

n=1

jz

n

j

2

< 1.Then the subset of H consisting of se-

quences fz

n

g with only ¯nitely many non-zero terms is a subspace but

not a closed subspace.

2.Orthogonal Decomposition

A vector x is orthogonal to a subset A of H if the inner product

hx;ai = 0 for every a 2 A.The set of all vectors orthogonal to A is

denoted by A

?

.

Exercise 2.Let A be a subset of H.

(1) Show that A

?

is a closed subspace.

1

2 MATH 650.THE RADON-NIKODYM THEOREM

(2) If A is a closed subspace,then show that (A

?

)

?

= A.

Proposition 1.If K is a closed subspace of H,if x is a vector and

if d = inffky ¡xk j y 2 Kg,then there exists a unique vector y

0

2 K

such that ky

0

¡xk = d.

Proof.Let y

n

be a sequence of vectors in K such that kx ¡y

n

k!d

as n!1.It follows from the parallelogram law (Exercise 3,x3.2)

applied to the vectors x ¡y

n

and x ¡y

m

that

ky

n

¡y

m

k

2

= 2kx ¡y

n

k +2kx ¡y

m

k

2

¡4kx ¡(y

n

+y

m

)=2k

2

for every n and m.Since (y

n

+y

m

)=2 2 K,it follows that

kx ¡(y

n

+y

m

)=2k

2

¸ d

2

and hence that

ky

n

¡y

m

k

2

· 2kx ¡y

n

k

2

+2kx ¡y

m

k

2

¡4d

2

:

As n!1 and m!1,the right side of this inequality tends to

2d

2

+ 2d

2

¡ 4d

2

= 0,so that fy

n

g is a Cauchy sequence,and so it

converges in H.If y

n

!y

0

,then y

0

2 K (because K is a closed

subspace and y

n

2 K) and,by the continuity of the norm (Theorem

14,x3.2),

ky

0

¡xk = lim

n

ky

n

¡xk = d:

If y

1

is another vector in K such that ky

1

¡xk = d,then (y

1

+y

0

)=2 2

K,so the de¯nition of d implies kx¡(y

0

+y

1

)=2k

2

¸ d

2

.This and the

parallelogram law give

ky

1

¡y

0

k

2

= 2kx ¡y

0

k

2

+2kx ¡y

1

k ¡4kx ¡(y

0

+y

1

)=2k

2

· 2d

2

+2d

2

¡4d

2

Thus ky

0

¡y

1

k = 0 and so y

0

= y

1

by the properties of the norm.¤

Theorem 1 (Othogonal Decomposition).Let K ½ H be a closed sub-

space.Then every vector x can be written in a unique way as a sum

x = Tx +Px,where Tx 2 K and Px?K.

Proof.Let Tx 2 K be the vector obtained in Proposition 1,and let

Px = x ¡ Tx.It has to be shown that Px?K,that is,that hx ¡

Tx;yi = 0 for all y in K.Let y 2 K with kyk = 1.For every complex

number ®,the vector Tx+®y 2 K because K is a subspace.Therefore,

for all ®,

kx ¡Txk

2

· kx ¡Tx ¡®yk

2

by the minimizing property of Tx.This inequality simpli¯es to

0 · j®j

2

¡

®hx ¡Tx;yi ¡®hy;x ¡Txi:

MATH 650.THE RADON-NIKODYM THEOREM 3

Taking in particular ® = hx¡Tx;yi,it obtains that 0 · ¡jhx¡Tx;yij

2

,

hence that x ¡Tx?K.

Uniqueness of the orthogonal decomposition is an easy exercise.¤

Corollary 1.Let K be a closed subspace such that K 6= H.Then

there exists a non-zero vector x 2 K

?

.

3.Riesz representation theorem

A linear functional on H is a map f:H!C such that

f(®x +¯y) = ®f(x) +¯f(y)

for every pair of vectors x and y and every pair of complex numbers ®

and ¯.

De¯nition 2.A continuous linear functional on H is a linear func-

tional f:H!C which is continuous,that is,whenever x

n

!x in H,

then f(x

n

)!f(x) in C.

Exercise 3.Let f:H!C be a linear functional and suppose that

there exists a constant C > 0 such that

jf(x)j · Ckxk

for every x 2 H.Show that f is continuous.

Example 3.Let H = L

2

(X;¹).Then

f 7!

Z

X

fd¹

is a linear functional.If the measure ¹(X) < 1,then it is also a

continuous linear functional.Indeed

¯

¯

¯

¯

Z

X

f ¢ ¹

¯

¯

¯

¯

· ¹(X)kfk

2

;

by the Schwarz inequality (Theorem 7,x3.2).

More generally,if z 2 H,then

x 2 H 7!hx;zi

is a continuous linear functional on H.It turns out that every contin-

uous linear functional is of this form.

Theorem 2 (Riesz Represenation).If f is a continuous linear func-

tional on H,then there exists a unique vector z 2 H such that

f(x) = hx;zi

for every x 2 H.

4 MATH 650.THE RADON-NIKODYM THEOREM

Proof.If f(x) = 0 for every x 2 H,then take z = 0.Assume thus f is

not identically 0.Let K = fx 2 H j f(x) = 0g.Then K is a subspace

of H,and is also closed because f is continuous and K = f

¡1

(0).

Because f 6´ 0,the closed subspace K 6= H.Thus,by Corollary 1,

there exits a non-zero vector y 2 H such that y?K.It may be

assumed that kyk = 1 (otherwise replace y by y=kyk).

Put u = (f(x))y ¡(f(y))x.Then u 2 K because f(u) = f(x)f(y) ¡

f(y)f(x) = 0.Therefore hu;yi = 0.But

hu;yi = h(f(x))y ¡(f(y))x;yi = f(x) ¡f(y)hx;yi

so that

f(x) = f(y)hx;yi:

Hence,f(x) = hx;zi with z =

f(y)y.

To see that z is unique,note that if hx;zi = hx;z

0

i for all x 2 H,

then u = z ¡z

0

is such that hx;ui = 0 for all x 2 H,hence u = 0.¤

4.The Lebesgue Decomposition Theorem

Let (X;F) be a measure space,and let ¹ and º be two measures on

X.

De¯nition 3.The measure º is said to be absolutely continuous with

respect to ¹,in symbols º Á ¹,if º(E) = 0 whenever ¹(E) = 0.

Example 4.If f is a non-negative ¹-integrable function on X,then

º(E) =

R

E

f ¢ ¹ de¯nes a measure on X which is absolutely continuous

with respect to ¹.

Exercise 4.Suppose that ¹ and º are ¯nite measures on (X;F) Then

º is absolutely continuous with respect to ¹ if and only if for every

"> 0 there exists ± > 0 such that º(E) <"for every E 2 F such that

¹(E) < ±.

De¯nition 4.The measures ¹ and º are singular,written ¹?º,if

there is an element E 2 F such that ¹(E) = 0 = º(X n E).

Example 5.Let X = R,F the Borel sets.Let ¹ be Lebesgue measure

and º be the measure given by º(E) = 1 if 0 2 E and º(E) = 0 if 0 =2 E.

Then ¹ and º are singular.

Theorem 3 (Lebesgue Decomposition).Let (X;F) be a measurable

space,and let ¹ and º be two ¯nite measures on X.Then there exist

unique measures º

a

and º

s

such that º = º

a

+º

s

,º

a

Á ¹ and º

s

?¹.

MATH 650.THE RADON-NIKODYM THEOREM 5

Proof.Let H be the Hilbert space H = L

2

(X;º +¹).Since the mea-

sures º and ¹ are ¯nite,so is º + ¹.Moreover,if f 2 H,then also

f 2 L

2

(X;º).Therefore

f 2 H 7!

Z

X

f ¢ º

is a continuous linear functional on H,because º · ¹ +º and Exam-

ple 3.By Theorem 2,there exists g 2 H such that

(y)

Z

f ¢ º =

Z

fg ¢ (º +¹);

for all f 2 H.Then

(¤)

Z

f(1 ¡g) ¢ (¹ +º) =

Z

f ¢ ¹:

This identity implies that g is real valued.Furthermore,g ¸ 0,for

otherwise,take f to be the characteristic function of the set fg(x) < 0g

for a contradiction.Likewise,g · 0.Thus it may be assumed that

0 · g(x) · 1 for all x.The monotone convergence theorem implies

that (¤) holds for all f ¸ 0.

Let A = fg = 1g and B = X n A.Then letting f = Â

A

in (¤) gives

¹(A) = 0.For E 2 F,set

º

s

(E):= º(E\A);and º

a

(E):= º(E\B)

Then º

s

and º

a

are measures,º = º

s

+º

a

,and º

s

?¹.

If E 2 F and ¹(E) = 0 and E ½ B,then

R

E

(1¡g)¢(¹+º) =

R

E

¹ = 0

by (¤) with (1¡g) > 0 on E,so (¹+º)(E) = 0 and º(E) = º

a

(E) = 0.

Hence º

a

Á ¹,and the existence part of the Lebesgue Decomposition

is proved.

To prove uniqueness,suppose that there is another decomposition

º = ½ +¾,with ½ Á ¹ and ¾?¹.Then ½(A) = 0 because ¹(A) = 0.

Thus for all E 2 F

º

s

(E) = º(E\A) = ¾(E\A) · ¾(E)

Thus º

s

· ¾ and ½ · º

a

.Then ¾ ¡ º

s

= º

a

¡ ½ is a measure both

absolutely continuous and singular with respect to ¹,so it is 0.Hence

½ = º

a

and ¾ = º

s

.¤

Example 6.Let X = R,let ¹ be Lebesgue measure on [0;2] and º be

Lebesgue measure on [1;3].Then º

a

is Lebesgue measure on [1;2] and

º

s

is Lebesgue measure on [2;3].

6 MATH 650.THE RADON-NIKODYM THEOREM

5.The Radon-Nikodym Theorem

Theorem4 (Radon-Nikodym).Let (X;F;¹) be a ¯nite measure space.

If º is a ¯nite measure on (X;F),absolutely continuous with respect to

¹,then there exists an integrable function h 2 L

1

(X;¹) such that

º(E) =

Z

E

h ¢ ¹

for all E 2 F.Any two such h are equal almost everywhere with respect

to ¹.

Proof.Note that the Lebesgue decomposition theoremgives º = º

a

+º

s

.

Since º Á ¹,it must be that º = º

a

and so º

s

= 0.Let h = g=(1 ¡g)

on B and h ´ 0 on A.If E 2 F,let f = hÂ

E

in (¤).Then,by (¤) and

(y),

Z

E

h ¢ ¹ =

Z

B\E

g ¢ (¹ +º) = º(B\E) = º(E):

To prove uniqueness,let k be another ¹-integrable function such

that º(E) =

R

E

k ¢ ¹,for all E 2 F.Then

R

E

(h ¡ k) ¢ ¹ = 0.Let

E

1

= fk < hg and E

2

= fk > hg.Integrating h ¡k over these sets it

obtains ¹(E

1

) = ¹(E

2

) = 0.Thus k = h ¹-almost everywhere.¤

The function h obtained in this theoremis called the Radon-Nikodym

derivative of º with respect to ¹,and it is usually denoted by dº=d¹.

The justi¯cation for this notation is that it satis¯es familiar calculus

properties.

Example 7.Let X = [0;1] and let f:X!X be a di®erentiable func-

tion whose derivative is bounded and nowhere zero.If ¹ is Lebesgue

measure on X,then º(E) = ¹(f

¡1

E) is a measure on X which is abso-

lutely continuous with respect to ¹,and the Radon-Nikodymderivative

dº=d¹ is jf

0

(x)j

6.Remarks and References

The proofs the Lebesgue Decomposition and Radon-Nikodym Theo-

rems given here,using the Riesz Representation Theorem as the main

tool,are due to J.von Neumann,see Dudley,Real Analysis and Prob-

ability,Chapman & Hall,New York,1989.

There are stronger versions of these theorems.For instance,in the

Lebesgue Decomposition it su±ces to assume that the measures are

¾-¯nite;in the Radon-Nikodym Theorem it su±ces that ¹ be ¾-¯nite

and º be ¯nite.See Dudley loc.cit.

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