10.13. RADON NIKODYM THEOREM 301

that ∪C is a countable set, {Fi}∞

i=1 . Otherwise, there would exist an infinite subset of ∪Cwith each set having measure less than − 1

n for some n ∈ N so λ would not be real valued.Letting F = ∪iFi, then E \F has no measurable subsets S for which λ (S) < 0 since, if itdid, C would not have been maximal. Thus E \F is positive. ■

A major result is the following, called a Hahn decomposition.

Theorem 10.13.5 Let λ be a signed measure on a measurable space (Ω,F ) . Thenthere are disjoint measurable sets P,N such that P is a positive set, N is a negative set, andP∪N = Ω.

Proof: If Ω is either positive or negative, there is nothing to show, so suppose Ω isneither positive nor negative. F will consist of collections of disjoint measurable sets Fsuch that λ (F)> 0. Thus each element of F is necessarily countable. Partially order F byset inclusion and use the Hausdorff maximal theorem to get C a maximal chain. Then, asin the above lemma, ∪C is countable, say {Pi}∞

i=1 because λ (F)> 0 for each F ∈ ∪C andλ has values in R. The sets in ∪C are disjoint because if A,B are two of them, then theyare both in a single element of C . Letting P ≡ ∪iPi, and N = PC, it follows from Lemma10.13.3 that P is positive. It is also the case that N must be negative because otherwise, Cwould not be maximal. ■

Clearly a Hahn decomposition is not unique. For example, you could have obtaineda different Hahn decomposition if you had considered disjoint negative sets F for whichλ (F)< 0 in the above argument .

Let k ∈ N,{

αkn}∞

n=0 be equally spaced points αkn = 2−kn. Then αk

2n = 2−k (2n) =2−(k−1)n≡ αk−1

n and αk+12n ≡ 2−(k+1)2n = αk

n. Similarly Nk+12n = Nk

n because these dependon the αk

n. Also let(Pk

n ,Nkn)

be a Hahn decomposition for the signed measure ν −αknµ

where ν ,µ are two finite measures. Now from the definition, Nkn+1 \Nk

n = Nkn+1∩Pk

n . Also,Nn ⊆Nn+1 for each n and we can take N0 = /0. Then

{Nk

n+1 \Nkn}∞

n=0 covers all of Ω exceptpossibly for a set of µ measure 0.

Lemma 10.13.6 Let S≡Ω\(∪nNk

n)= Ω\

(∪nNl

n)

for any l. Then µ (S) = 0.

Proof: S = ∩nPkn so for all n,ν (S)−αk

nµ (S) ≥ 0. But letting n→ ∞, it must be thatµ (S) = 0. ■

As just noted, if E ⊆ Nkn+1 \Nk

n , then

ν (E)−αknµ (E)≥ 0≥ ν (E)−α

kn+1µ (E) , so α

kn+1µ (E)≥ ν (E)≥ α

knµ (E) (10.16)

Nkn

Nkn+1

αkn+1µ(E)≥ ν(E)≥ αk

nµ(E)

Then define f k (ω)≡ ∑∞n=0 αk

nX∆kn(ω) where ∆k

m ≡ Nkm+1 \Nk

m. Thus,

f k =∞

∑n=0

αk+12n X(Nk+1

2n+2\Nk+12n ) =

∞

∑n=0

αk+12n X

∆k+12n+1

+∞

∑n=0

αk+12n X

∆k+12n

≤∞

∑n=0

αk+12n+1X∆

k+12n+1

+∞

∑n=0

αk+12n X

∆k+12n

= f k+1 (10.17)