31.3. DIFFERENTIATION OF RADON MEASURES 1093

31.3.1 Radon Nikodym Theorem for Radon MeasuresThe Radon Nikodym theorem is an abstract result but this will be a special version forRadon measures which is based on these covering theorems and related theory.

Definition 31.3.4 Let λ ,µ be two Radon measures defined on F , a σ algebra of subsetsof an open set U. Then λ ≪ µ means that whenever µ (E) = 0, it follows that λ (E) = 0.

Next is a representation theorem for λ in terms of an integral involving Dµ λ .

Theorem 31.3.5 Let λ and µ be Radon measures defined on F a σ algebra of the openset U then there exists a set of µ measure zero N such that Dµ λ (x) exists off N and ifE ⊆ NC,E ∈F , then

λ (E) =∫

U

(Dµ λ

)XEdµ.

If λ ≪ µ,λ (E) =∫

E Dµ λdµ . In any case, λ (E)≥∫

E Dµ λdµ .

Proof: The proof is based on Lemma 31.3.2. Let E ⊆ NC where N has measure 0and includes the set Z along with the set where the symmetric derivative does not exist.It can be assumed that N is a Gδ set. Assume E is bounded to begin with. Then E ∩{

x ∈ NC : Dµ λ (x) = 0}

has measure zero. This is because by Lemma 31.3.2,

λ(E ∩

{x ∈ NC : Dµ λ (x) = 0

})≤ aµ

(E ∩

{x ∈ NC : Dµ λ (x) = 0

})≤ aµ (E) , µ (E)< ∞

for all positive a and so

λ(E ∩

{x ∈ NC : Dµ λ (x) = 0

})= 0

Thus, the set where Dµ λ (x) = 0 can be ignored.Let{

ank

}∞

k=1 be positive numbers such that∣∣an

k−ank+1

∣∣= 2−n. Specifically, let{

ank

}∞

k=0be given by

0,2−n,2(2−n) ,3(2−n) ,4(2−n) , ...

Define disjoint half open intervals whose union is all of (0,∞) , Ink having end points an

k−1and an

k . Say Ink = (an

k−1,ank ].

Enk ≡ E ∩

{x ∈ Rp : Dµ λ (x) ∈ In

k}≡ E ∩

(Dµ λ

)−1(In

k )

Since the intervals are Borel sets,(Dµ λ

)−1 (Ink

)is measurable. Thus ∪∞

k=1Enk = E and the

k→ Enk are disjoint measurable sets. From Lemma 31.3.2,

µ (Enk )an

k ≥ λ (Enk )≥ an

k−1µ (Enk )

Then∞

∑k=1

ank µ (En

k )≥ λ (E) =∞

∑k=1

λ (Enk )≥

∞

∑k=1

ank−1µ (En

k )