MATH 650.THE RADONNIKODYM THEOREM
This note presents two important theorems in Measure Theory,the
Lebesgue Decomposition and RadonNikodym 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 nonzero 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 RADONNIKODYM 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 RADONNIKODYM 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 nonzero 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 RADONNIKODYM 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 nonzero 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 nonnegative ¹integrable 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 RADONNIKODYM 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 RADONNIKODYM THEOREM
5.The RadonNikodym Theorem
Theorem4 (RadonNikodym).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 ¹integrable 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 ¹almost everywhere.¤
The function h obtained in this theoremis called the RadonNikodym
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 RadonNikodymderivative
dº=d¹ is jf
0
(x)j
6.Remarks and References
The proofs the Lebesgue Decomposition and RadonNikodym 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 RadonNikodym Theorem it su±ces that ¹ be ¾¯nite
and º be ¯nite.See Dudley loc.cit.
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