1094 CHAPTER 31. DIFFERENTIATION, RADON MEASURES

Let ln (x) ≡ ∑∞k=1 an

k−1X(Dµ λ)−1(In

k )(x) and un (x) ≡ ∑

∞k=1 an

kX(Dµ λ)−1(In

k )(x) . Then the

above implies

λ (E) ∈[∫

Elndµ,

∫E

undµ

]Now both ln and un converge to Dµ λ (x) which is nonnegative and measurable as shownearlier. The construction shows that ln increases to Dµ λ (x) . Also, un (x)− ln (x) = 2−n.Thus

λ (E) ∈[∫

Elndµ,

∫E

lndµ +2−nµ (E)

]By the monotone convergence theorem, this shows λ (E) =

∫E Dµ λdµ.

Now if E is an arbitrary set in NC, maybe not bounded, the above shows

λ (E ∩B(0,n)) =∫

E∩B(0,n)Dµ λdµ

Let n→ ∞ and use the monotone convergence theorem. Thus for all E ⊆ NC, λ (E) =∫E Dµ λdµ . For the last claim,

∫E Dµ λdµ =

∫E∩NC Dµ λdµ = λ

(E ∩NC

)≤ λ (E).

In case, λ ≪ µ, it does not matter that E ⊆ NC because, since µ (N) = 0, so is λ (N)and so

λ (E) = λ(E ∩NC)= ∫

E∩NCDµ λdµ =

∫E

Dµ λdµ

for any E ∈F .What if λ and µ are just two arbitrary Radon measures defined on F ? What then? It

was shown above that Dµ λ (x) exists for µ a.e. x, off a Gδ set N of µ measure 0 whichincludes Z, the set of x where µ (B(x,r)) = 0 for some r > 0. Also, it was shown abovethat if E ⊆ NC, then λ (E) =

∫E Dµ λ (x)dµ. Define for arbitrary E ∈F ,

λ µ (E)≡ λ(E ∩NC) , λ⊥ (E)≡ λ (E ∩N)

Then

λ (E) = λ (E ∩N)+λ(E ∩NC)= λ⊥ (E)+λ µ (E)

= λ (E ∩N)+∫

E∩NCDµ λ (x)dµ

= λ (E ∩N)+∫

EDµ λ (x)dµ ≡ λ (E ∩N)+λ µ (E)

≡ λ⊥ (E)+λ µ (E)

This shows most of the following corollary.

Corollary 31.3.6 Let µ,λ be two Radon measures. Then there exist two measures, λ µ ,λ⊥such that

λ µ ≪ µ, λ = λ µ +λ⊥

and a set of µ measure zero N such that

λ⊥ (E) = λ (E ∩N)