20.1. RADON NIKODYM THEOREM 599

From 20.1.1, whenever g ∈ L2(Ω,µ +λ ),∫Ω

g(1−h)dλ =∫

Ω

hgdµ . (20.1.3)

Now let E be a measurable set and define

g(x)≡n

∑i=0

hi(x)XE(x)

in 20.1.3. This yields ∫E(1−hn+1(x))dλ =

∫E

n+1

∑i=1

hi(x)dµ . (20.1.4)

Let f (x) = ∑∞i=1 hi(x) and use the Monotone Convergence theorem in 20.1.4 to let n→ ∞

and concludeλ (E) =

∫E

f dµ .

f ∈ L1(Ω,µ) because λ is finite.The function, f is unique µ a.e. because, if g is another function which also serves to

represent λ , consider for each n ∈ N the set,

En ≡[

f −g >1n

]and conclude that

0 =∫

En

( f −g)dµ ≥ 1n

µ (En) .

Therefore, µ (En) = 0. It follows that

µ ([ f −g > 0])≤∞

∑n=1

µ (En) = 0

Similarly, the set where g is larger than f has measure zero. This proves the theorem.Case where it is not necessarily true that λ ≪ µ.In this case, let N = [h≥ 1] and let g = XN . Then

λ (N) =∫

Nh d(µ +λ )≥ µ(N)+λ (N).

and so µ (N) = 0. Now define a measure, λ⊥ by

λ⊥ (E)≡ λ (E ∩N)

so λ⊥ (E ∩N) = λ (E ∩N∩N)≡ λ⊥ (E) and let λ || ≡ λ −λ⊥. Therefore,

µ (E) = µ(E ∩NC)