The Radon Nikodym theorem is an abstract result but this will be a special version for Radon measures
which is based on these covering theorems and related theory.
Definition 9.5.1Let λ,μ be two Radon measures defined on ℱ. Then λ ≪ μ means that wheneverμ
(E )
= 0, it follows that λ
(E)
= 0.
Recall that the symmetric derivative exists for μ a.e. x. Then the absolute continuity assumed above
indicates that this happens also for λ a.e. x.
Theorem 9.5.2Let λ and μ be Radonmeasures and suppose λ ≪ μ. Then for all E a μ measurableset,
∫
λ (E ) = (Dμλ)dμ.
E
In addition to this, for λ,μ arbitrary Radon measures, there exists a set of μ measure zero N such thatD_{μ}λ
(x)
exists off N and if E ⊆ N^{C}, then
∫
λ (E ) = (Dμλ)dμ.
E
Proof:Let t > 1 and let E be a μ measurable set which is bounded and a subset of N^{C} where N is
the exceptional set of μ measure zero in Theorem 9.4.3 off of which μ
is finite due to the assumption that E is bounded and μ is a Radon measure.
Therefore, by Lemma 9.4.2,
∞ −1
λ (E ) = ∑ λ(Em )+ ∑ λ(Em )
m=0 h=−∞
∞∑ m+1 ∑−1 m ∑∞ m −∑1 m
≤ t μ(Em )+ t μ (Em) = t t μ (Em) + t μ(Em )
m=0 h=−∞ m=0 h= −∞
∞∑ m −1 ∑−1 m+1
= t t μ (Em )+ t t μ(Em )
m=0 h=− ∞
∞∑ ∫ −∑1 ∫
≤ t D μλ(x)dμ +t−1 D μλ(x)dμ
m=0 Em h= −∞ Em
∫ ∫
= t Dμλ (x)dμ+ t−1 D μλ(x)dμ
E+ E−
where E_{−} is the union of the E_{m} for m < 0. Thus E_{−} is the set of x in E such that D_{μ}λ
(x)
< 1 and E_{+} is
the union of the E_{m} for m ≥ 0 so it is the set of x in E out of the exceptional set where D_{μ}λ
(x)
≥ 1.
Hence E_{−} and E_{+} do not depend on m. Also by this same lemma,
∑∞ −∑ 1 ∞∑ m −∑1 m+1
λ(E ) = λ(Em )+ λ (Em ) ≥ t μ (Em )+ t μ(Em )
m=0 m=−∞ m=0 m= −∞
∞∑ ∑−1
≥ t−1 tm+1μ(Em )+ t tm μ(Em )
m=0 m= −∞
−1 ∞∑ ∫ −∑1 ∫
≥ t Dμλ (x)dμ+ t Dμλ (x )dμ
m=0 Em m= −∞ Em
−1∫ ∫
= t Dμλ (x)dμ+ t D μλ(x)dμ
E+ E−
Thus,
∫ −1∫
t E Dμ λ(x)dμ+ t E D μλ(x)dμ
+ ∫ − ∫
≥ λ (E ) ≥ t−1 Dμ λ(x)dμ+ t D μλ(x)dμ
E+ E −
and letting t → 1, it follows
∫
λ (E) = D λ(x)dμ. (9.16)
E μ
(9.16)
Now if E is an arbitrary measurable set, contained in N^{C}, this formula holds with E replaced with
E ∩B
(0,k)
. Letting k →∞ and using the monotone convergence theorem, the above formula holds for all
E ⊆ N^{C}. No assumption of absolute continuity has been used up till now. It is in going from E ⊆ N^{C} to
arbitrary E where this assumption is used.
Assume now that λ ≪ μ. Since N is a set of μ measure zero, it follows N is also a set of λ measure zero
due to the assumption of absolute continuity. Therefore 9.16 continues to hold for arbitrary μ measurable
sets, E even if they are not contained in N^{C}. ■
What if λ and μ are just two arbitrary Radon measures defined on ℱ? What then? It was shown above
that D_{μ}λ
(x)
exists for μ a.e. x. Also, it was shown above that if E ⊆ N^{C}, then
∫
λ(E) = ED μλ(x)dμ
Define for arbitrary E ∈ℱ,
∫
λ (E ) ≡ D λ (x)dμ
μ E μ
Then you could let
λ⊥(E ) ≡ λ(E )− λμ(E)
Letting N be the set on which the derivative D_{μ}λ
(x)
does not exist, it was shown above that μ
(N )
= 0.
Then
λ (E ) = λ (E ∩ N )+ λ(E ∩ NC )
λ⊥ (E)+ λμ (E ) = λ(E ) = λ(E ∩ N) + λ(E ∩N C)
∫
= λ(E ∩ N )+ Dμλ (x )dμ
∫E∩NC
= λ(E ∩ N )+ D μλ(x)dμ ≡ λ(E ∩ N) + λμ(E)
E
and this shows that λ_{⊥}
(E)
= λ
(E ∩N )
. This shows most of the following corollary.
Corollary 9.5.3Let μ,λ be two Radon measures. Then there exist two measures, λ_{μ},λ_{⊥}suchthat
λμ ≪ μ, λ = λ μ + λ⊥
and a set of μ measure zero N such that
λ⊥(E ) = λ(E ∩ N)
Also λ_{μ}is given by the formula
∫
λ (E ) = D λ (x)dμ
μ E μ
Proof:It only remains to verify that λ_{μ} given above satisfies λ_{μ}≪ μ. However, this is obvious because
if μ