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 18.5.1Let λ,μ be two Radon measures defined on ℱ. Then λ ≪ μmeans that whenever μ
(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 18.5.2Let λ and μ be Radonmeasures and suppose λ ≪ μ. Then for allE a μ measurable set,
∫
λ(E ) = E (D μλ)dμ.
In addition to this, for λ,μ arbitrary Radon measures, there exists a set of μ measure zero Nsuch that D_{μ}λ
(x)
exists off N and if E ⊆ N^{C}, then
∫
λ(E ) = E (D μλ)dμ.
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 18.4.3 off of which
μ
∑∞ ∫ ∑−1 ∫
≥ t−1 Dμλ (x )dμ+ t Dμλ (x) dμ
m=0 Em m= −∞ Em
∫ ∫
−1
= t E+ Dμλ (x )dμ+ t E− Dμλ (x)dμ
Thus,
∫ ∫
−1
t E+ Dμλ (x)dμ+ t E− D μλ(x)dμ
∫ ∫
≥ λ (E ) ≥ t−1 Dμλ (x)dμ+ t D μλ(x)dμ
E+ E−
and letting t → 1, it follows
∫
λ (E ) = Dμ λ(x)dμ. (18.16)
E
(18.16)
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 18.16 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,
∫
λμ(E ) ≡ E Dμλ (x )dμ
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 )
( C)
λ⊥ (E )+ λμ (E ) = λ(E) = λ(E∫∩ N)+ λ E ∩N
= λ(E ∩ N) + Dμλ (x) dμ
∫E ∩NC
= λ(E ∩ N) + E D μλ(x)dμ ≡ λ(E ∩ N)+ λμ (E)
and this shows that λ_{⊥}
(E)
= λ
(E ∩N )
. This shows most of the following corollary.
Corollary 18.5.3Let μ,λ be two Radon measures. Then there exist two measures,λ_{μ},λ_{⊥}such that
λμ ≪ μ, λ = λμ + 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