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 23.1.14.
Definition 23.2.1Let μ and λ be two measures defined on a σ-algebra S, of subsets of a set, Ω.λ is absolutely continuous with respect to μ,written as λ ≪ μ, if λ(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 23.2.2(Radon Nikodym) Let λ and μ be finite measures defined on a σ-algebra, S, ofsubsets of Ω. Suppose λ ≪ μ. Then there exists a unique f ∈ L^{1}(Ω,μ) such that 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 E measurable,λ_{⊥}
(E)
= λ
(E ∩N )
= λ_{⊥}
(E ∩ N)
. In this case the two measures, λ_{⊥}and λ_{||}are unique andtherepresentation of λ = λ_{⊥} + λ_{||}is called the Lebesgue decomposition of λ. The measure λ_{||}is the absolutely continuous part of λ and λ_{⊥}is called the singular part of λ. In words, λ_{⊥}is “supported” on a set of μ measure zero while λ_{||}is supported on the complement of thisset.
_{2} is the L^{2} norm of g taken with respect to μ + λ. Therefore, since Λ is bounded, it follows from
Theorem 23.1.13 on Page 1721 that Λ ∈ (L^{2}(Ω,μ + λ))^{′}, the dual space L^{2}(Ω,μ + λ). By the Riesz
representation theorem in Hilbert space, Theorem 23.1.14, there exists a unique h ∈ L^{2}(Ω,μ + λ)
with
∫ ∫
Λg = gdλ = hgd(μ+ λ). (23.9)
Ω Ω
(23.9)
The plan is to show h is real and nonnegative at least a.e. Therefore, consider the set where Imh is
positive.
(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
which is a contradiction because of the strict inequality which results if λ_{||}
(E )
> 0. 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
ˆN
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}≡
[ ]
fˆ− f > 1∕k
for k ∈ ℕ. Then on observing that
λ_{⊥}−
ˆ
λ
_{⊥} =
ˆ
λ
_{||}− λ_{||}
( ) ( ) ∫ ( )
0 = λ − ˆλ E ∩(N ∪ N)C = fˆ− f dμ
⊥ ⊥ k 1 Ek∩(N1∪N)C
1- ( C ) -1
≥ k μ Ek ∩ (N1 ∪ N ) = k μ(Ek).
and so μ
(Ek )
= 0. The last equality follows from
( ) ( ) ( )
μ Ek ∩ (N1 ∪ N )C = μ Ek ∩ N C1 ∩N C2 = μ Ek ∩ NC1 = μ(Ek)
Therefore, μ
([ˆf − f > 0])
= 0 because
[fˆ− f > 0]
= ∪_{k=1}^{∞}E_{k}. It follows
fˆ
≤ f μ a.e. Similarly,
ˆf
≥ fμ 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 23.2.3Suppose λ ≪ μ and there exist sets S_{n}∈S, the σ algebra of measurable setswith
∞
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 23.2.2, 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
∞ ∫ N ∫
∑ XE ∩S (x)f(x)dμ = lim ∑ XE ∩S (x)f (x)dμ
n=1 n N →∞ n=1 n
∫ ∑N
= lim XE ∩Sn(x)f (x)dμ
N →∞ n=1
∫ ∑∞ ∫
= XE∩Sn(x)f(x)dμ = fdμ.
n=1 E
This proves the existence part of the corollary.
To see f is unique, suppose f_{1} and f_{2} both work and consider for n ∈ ℕ
[ 1]
Ek ≡ f1 − f2 >--.
k
Then
∫
0 = λ(Ek ∩Sn )− λ(Ek ∩ Sn) = E ∩S f1(x) − f2(x)dμ.
k n
Hence μ(E_{k}∩ S_{n}) = 0 for all n so
μ(Ek) = lim μ(E ∩Sn) = 0.
n→ ∞
Hence μ(
[f1 − f2 > 0]
) ≤∑_{k=1}^{∞}μ
(Ek)
= 0. Therefore, λ
([f1 − f2 > 0])
= 0 also. Similarly
(μ+ λ)([f1 − f2 < 0]) = 0.■
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 [17]. 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
23.2.2.
Corollary 23.2.4Let
(Ω,S)
be a set with a σ algebra of sets. Suppose λ and μ are two measures definedon the sets of S and suppose there exists a sequence of disjoint sets of S,
{Ωi}
_{i=1}^{∞}such thatλ
(Ωi)
,μ
(Ωi)
< ∞,Ω = ∪_{i}Ω_{i}. Then there is a set of μ measure zero, N and measures λ_{⊥}and λ_{||}suchthat
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 23.2.2 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
= ∑ ∑ λi(Ej ∩Ωi) = ∑ ∑ λ(Ej ∩Ωi ∩Ni )
j i ⊥ j i
∑ ∑ i ∑
= λ⊥(Ej ∩Ωi) = λ⊥(Ej).
j i j
The argument for λ_{||} is similar. Now
∑ ∑ i
μ (N ) = μ (N ∩Ωi) = μ (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}.■