# MATH 650. THE RADON-NIKODYM THEOREM This note presents ...

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Oct 8, 2013 (4 years and 7 months ago)

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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) 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:
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.
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

s

a
Á ¹ and º
s
?¹.
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

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].
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

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.