344 CHAPTER 11. REGULAR MEASURES

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)

Thenλ (E) = λ (E ∩N)+λ

(E ∩NC)= λ⊥ (E)+λ µ (E)

= λ (E ∩N)+∫

E∩NCDµ λ (x)dµ = λ (E ∩N)+

∫E

Dµ λ (x)dµ

≡ λ (E ∩N)+λ µ (E)≡ λ⊥ (E)+λ µ (E)

This shows the following corollary.

Corollary 11.13.3 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) . Also λ µ is given by the formula λ µ (E)≡

∫E Dµ λ (x)dµ.

Proof: If x ∈ N, this could happen two ways, either x ∈ Z or Dµ λ (x) fails to exist.It only remains to verify that λ µ given above satisfies λ µ ≪ µ. However, this is obviousbecause if µ (E) = 0, then

∫E Dµ λ (x)dµ = 0. ■

Since Dµ λ (x)=Dµ λ (x)XNC (x) , it doesn’t matter which we use but maybe Dµ λ (x)doesn’t exist at some points of N, so although I will use Dµ λ (x) , it might be more preciseto use Dµ λ (x)XNC (x).

This is sometimes called the Lebesgue decomposition.How does this relate to Corollary 10.13.14? It tells how to find the function f in that

Corollary as a symmetric derivative. This is very useful when you want to have an explicitdescription of the Radon Nikodym derivative.

11.14 Absolutely Continuous FunctionsCan you integrate the derivative to get the function as in calculus? The answer is that some-times you can and when this is the case, the function is called absolutely continuous. This isexplained in this section. Recall the following which summarizes Theorems 9.9.1 on Page257 and 9.7.4 on Page 250. In what follows m will be one dimensional Lebesgue measure.Recall that for F increasing, F (x+)≡ limh→0+ F (x+h) ,F (x−)≡ limh→0+ F (x−h) .

Theorem 11.14.1 Let F be an increasing function on R. Then there is an outermeasure µ and a σ algebra F on which µ is a measure such that F contains the Borelsets. This measure µ satisfies

µ ([a,b]) = F (b+)−F (a−) , µ ((a,b)) = F (b−)−F (a+)