This chapter is on various representation theorems. The first theorem, the Radon Nikodym
Theorem, is a representation theorem for one measure in terms of another. The approach
given here is due to Von Neumann and depends on the Riesz representation theorem for
Hilbert space, Theorem 24.1.14 on Page 1702.
Definition 25.1.1Let μ and λ be two measures defined on a σ-algebra S, ofsubsets of a set, Ω. λ is absolutely continuous with respect to μ,written as λ ≪ μ, ifλ(E) = 0 whenever μ(E) = 0.
It is not hard to think of examples which should be like this. For example, suppose one
measure is volume and the other is mass. If the volume of something is zero, it is reasonable
to expect the mass of it should also be equal to zero. In this case, there is a function called
the density which is integrated over volume to obtain mass. The Radon Nikodym theorem is
an abstract version of this notion. Essentially, it gives the existence of the density
function.
Theorem 25.1.2(Radon Nikodym) Let λ and μ be finite measures defined on aσ-algebra, S, of subsets of Ω. Suppose λ ≪ μ. Then there exists a unique f ∈ L^{1}(Ω,μ) suchthat f(x) ≥ 0 and
∫
λ(E ) = fdμ.
E
If it is not necessarily the case that λ ≪ μ, there are two measures, λ_{⊥}and λ_{||}such thatλ = λ_{⊥} + λ_{||},λ_{||}≪ μ and there exists a set of μ measure zero, N such that for all Emeasurable, λ_{⊥}
(E)
= λ
(E ∩ N)
= λ_{⊥}
(E ∩ N)
. In this case the two measures, λ_{⊥}and λ_{||}areunique and therepresentation of λ = λ_{⊥} + λ_{||}is called the Lebesgue decomposition of λ. Themeasure λ_{||}is the absolutely continuous part of λ and λ_{⊥}is called the singular part ofλ.
_{2} is the L^{2} norm of g taken with respect to μ + λ. Therefore, since Λ is bounded, it
follows from Theorem 23.1.4 on Page 1590 that Λ ∈ (L^{2}(Ω,μ + λ))^{′}, the dual space
L^{2}(Ω,μ + λ). By the Riesz representation theorem in Hilbert space, Theorem 24.1.14, there
exists a unique h ∈ L^{2}(Ω,μ + λ) with
∫ ∫
Λg = gdλ = hgd(μ+ λ). (25.1)
Ω Ω
(25.1)
The plan is to show h is real and nonnegative at least a.e. Therefore, consider the set where
Imh is positive.
∫ ∫
0 = (Im h)d(μ+ λ) ≥ (Im h)d(μ+ λ) ≥ 1(μ + λ)(En)
E En n
where
{ 1}
En ≡ x : Im h(x) ≥ n
Thus
(μ + λ)
(E_{n}) = 0 and since E = ∪_{n=1}^{∞}E_{n}, it follows
(μ + λ)
(E) = 0. A similar
argument shows that for
E = {x ∈ Ω : Im h (x) < 0},
(μ + λ)(E) = 0. Thus there is no loss of generality in assuming h is real-valued.
The next task is to show h is nonnegative. This is done in the same manner
as above. Define the set where it is negative and then show this set has measure
zero.
Let E ≡{x : h(x) < 0} and let E_{n}≡{x : h(x) < −
-1
n
}. Then let g = X_{En}. Since
E = ∪_{n}E_{n}, it follows that if
∫ ∫
n+1 n∑+1 i
E (1 − h (x ))dλ = E h(x)dμ. (25.4)
i=1
(25.4)
Let f(x) = ∑_{i=1}^{∞}h^{i}(x) and use the Monotone Convergence theorem in 25.4 to let n →∞
and conclude
∫
λ(E ) = fdμ.
E
f ∈ L^{1}(Ω,μ) because λ is finite.
The function, f is unique μ a.e. because, if g is another function which also serves to
represent λ, consider for each n ∈ ℕ the set,
[ ]
E ≡ f − g > 1
n n
and conclude that
∫
0 = (f − g)dμ ≥ 1-μ(E ).
En n n
Therefore, μ
(En)
= 0. It follows that
∞
μ([f − g > 0]) ≤ ∑ μ (E ) = 0
n=1 n
Similarly, the set where g is larger than f has measure zero. This proves the theorem when
μ ≫ λ.
Case where it is not necessarily true that λ ≪ μ.
In this case, let N =
[h ≥ 1]
and let g = X_{N}. Then
∫
λ(N ) = N hd(μ + λ) ≥ μ(N )+ λ(N ).
and so μ
(N )
= 0. Now define measures, λ_{⊥}, λ_{||}
( C)
λ⊥ (E ) ≡ λ (E ∩ N ), λ||(E ) ≡ λ E ∩N so λ = λ⊥ + λ||
Since μ
(N )
= 0,
( )
μ (E) = μ E ∩N C
Suppose then that μ
(E )
= μ
( )
E ∩ N C
= 0. Does λ_{||}
(E )
= 0? Then since h < 1 on N^{C}, if
λ_{||}
(E )
> 0,
∫
λ||(E) ≡ λ (E ∩N C) = hd(μ+ λ)
( ) E(∩NC )
< μ E ∩N C + λ E ∩ NC = μ(E )+ λ||(E) ,
which is a contradiction. Therefore, λ_{||}≪ μ because if μ
(E )
= 0, then λ_{||}
(E)
= 0.
It only remains to verify the two measures λ_{⊥} and λ_{||} are unique. Suppose then that
ˆ
λ
_{⊥}
and
ˆ
λ
_{||} play the roles of λ_{⊥} and λ_{||} respectively. Let
ˆ
N
play the role of N in the
definition of
ˆλ
_{⊥} and let
ˆf
play the role of f for
ˆλ
_{||}. I will show that f =
ˆf
μ a.e.
Let E_{k}≡
[ ]
ˆf − f > 1∕k
for k ∈ ℕ. Then on observing that λ_{⊥}−
ˆλ
_{⊥} =
ˆλ
_{||}− λ_{||}
( ) ( ) ∫ ( )
0 = λ⊥ − ˆλ⊥ Ek ∩ (N1 ∪N )C = C ˆf − f dμ
( ) Ek∩(N1∪N )
≥ 1μ Ek ∩(N1 ∪ N)C = 1-μ(Ek).
k k
and so μ
(Ek )
= 0. Therefore, μ
([ ])
ˆf − f > 0
= 0 because
[ ]
fˆ− f > 0
= ∪_{k=1}^{∞}E_{k}. It
follows
ˆf
≤ f μ a.e. Similarly,
ˆf
≥ f μ a.e. Therefore,
ˆλ
_{||} = λ_{||} and so λ_{⊥} =
ˆλ
_{⊥} also.
■
The f in the theorem for the absolutely continuous case is sometimes denoted by
ddλμ
and
is called the Radon Nikodym derivative.
The next corollary is a useful generalization to σ finite measure spaces.
Corollary 25.1.3Suppose λ ≪ μ and there exist sets S_{n}∈S with
Sn ∩ Sm = ∅, ∪∞n=1 Sn = Ω,
and λ(S_{n}), μ(S_{n}) < ∞. Then there exists f ≥ 0, where f is μ measurable, and
∫
λ(E) = fdμ
E
for all E ∈S. The function f is μ + λa.e. unique.
Proof: Define the σ algebra of subsets of S_{n},
Sn ≡ {E ∩Sn : E ∈ S}.
Then both λ, and μ are finite measures on S_{n}, and λ ≪ μ. Thus, by Theorem 25.1.2, there
exists a nonnegative S_{n} measurable function f_{n},with λ(E) = ∫_{E}f_{n}dμ for all E ∈S_{n}. Define
f(x) = f_{n}(x) for x ∈ S_{n}. Since the S_{n} are disjoint and their union is all of Ω, this defines f
on all of Ω. The function, f is measurable because
This version of the Radon Nikodym theorem will suffice for most applications, but more
general versions are available. To see one of these, one can read the treatment in Hewitt and
Stromberg [15]. This involves the notion of decomposable measure spaces, a generalization of
σ finite.
Not surprisingly, there is a simple generalization of the Lebesgue decomposition part of
Theorem 25.1.2.
Corollary 25.1.4Let
(Ω,S)
be a set with a σ algebra of sets. Suppose λ and μ are twomeasures defined on the sets of S and suppose there exists a sequence of disjoint sets of S,
{Ωi}
_{i=1}^{∞}such that λ
(Ωi)
,μ
(Ωi)
< ∞,Ω = ∪_{i}Ω_{i}. Then there is a set of μ measure zero, Nand measures λ_{⊥}and λ_{||}such that
λ⊥ +λ || = λ,λ|| ≪ μ, λ⊥(E ) = λ(E ∩ N ) = λ⊥ (E ∩ N ).
Proof: Let S_{i}≡
{E ∩Ωi : E ∈ S}
and for E ∈S_{i}, let λ^{i}
(E)
= λ
(E)
and μ^{i}
(E )
= μ
(E )
.
Then by Theorem 25.1.2 there exist unique measures λ_{⊥}^{i} and λ_{||}^{i} such that λ^{i} = λ_{⊥}^{i} + λ_{||}^{i}, a
set of μ^{i} measure zero, N_{i}∈S_{i} such that for all E ∈S_{i}, λ_{⊥}^{i}
(E)
= λ^{i}
(E ∩ Ni)
and λ_{||}^{i}≪ μ^{i}.
Define for E ∈S
∑ ∑
λ⊥ (E ) ≡ λi⊥ (E ∩Ωi),λ||(E ) ≡ λi||(E ∩ Ωi),N ≡ ∪iNi.
i i
First observe that λ_{⊥} and λ_{||} are measures.
∑ ∑ ∑
λ⊥ (∪∞j=1Ej) ≡ λi⊥ (∪∞j=1Ej ∩Ωi) = λi⊥ (Ej ∩ Ωi)
i i j
∑ ∑ i ∑ ∑
= λ ⊥(Ej ∩Ωi) = λ(Ej ∩ Ωi ∩Ni)
∑j ∑i ∑j i
= λi⊥(Ej ∩Ωi) = λ⊥(Ej).
j i j
The argument for λ_{||} is similar. Now
∑ ∑
μ(N ) = μ (N ∩Ωi) = μi(Ni ) = 0
i i
and
∑ i ∑ i
λ⊥ (E) ≡ λ⊥ (E ∩Ωi) = λ (E ∩Ωi ∩Ni )
∑i i
= λ (E ∩ Ωi ∩N ) = λ (E ∩ N ).
i
Also if μ
(E )
= 0, then μ^{i}
(E ∩Ωi)
= 0 and so λ_{||}^{i}
(E ∩ Ωi)
= 0. Therefore,
∑ i
λ||(E) = λ||(E ∩Ωi) = 0.
i
The decomposition is unique because of the uniqueness of the λ_{||}^{i} and λ_{⊥}^{i} and the
observation that some other decomposition must coincide with the given one on the
Ω_{i}.■