Definition 14.1.1Let 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 14.1.2(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 and therepresentation of λ = λ_{⊥} + λ_{||}is called the Lebesgue decomposition of λ. The measure λ_{||}is the absolutelycontinuous part of λ and λ_{⊥}is called the singular part of λ.
_{2} is the L^{2} norm of g taken with respect to μ + λ. Therefore, since Λ is bounded, it follows from
Theorem 13.1.8 on Page 1010 that Λ ∈ (L^{2}(Ω,μ + λ))^{′}, the dual space L^{2}(Ω,μ + λ). By the Riesz
representation theorem in Hilbert space, Theorem 13.2.14, there exists a unique h ∈ L^{2}(Ω,μ + λ)
with
∫ ∫
Λg = gdλ = hgd(μ+ λ). (14.1)
Ω Ω
(14.1)
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. 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. Therefore, μ
([ ])
ˆf − f > 0
= 0 because
[ ]
fˆ− f > 0
= ∪_{k=1}^{∞}E_{k}. It follows
fˆ
≤ f μ
a.e. Similarly,
ˆ
f
≥ f μ a.e. Therefore,
ˆ
λ
_{||} = λ_{||} and so λ_{⊥} =
ˆ
λ
_{⊥} also. ■
The f in the theorem for the absolutely continuous case is sometimes denoted by
dλ
dμ
and is called the
Radon Nikodym derivative.
The next corollary is a useful generalization to σ finite measure spaces.
Corollary 14.1.3Suppose 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) = Efdμ
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 14.1.2, 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 [62]. This involves the notion of decomposable measure spaces, a generalization of σ
finite.