31.2. SLICING MEASURES 1085

If E ∈Sν not necessarily bounded, let Em = E ∩B(0,m) and then Em ∈Sµ and µ (Em) =ν (Em) . Letting m→ ∞,E ∈ Sµ and µ (E) = ν (E) . Similarly, Sµ ⊆ Sν and the twomeasures are equal on Sµ .

The main result in the section is the following theorem.

Theorem 31.2.3 Let µ be a finite Radon measure on Rn+m defined on a σ algebra, F .Then there exists a unique finite Radon measure α, defined on a σ algebra S , of sets ofRn which satisfies

α (E) = µ (E×Rm) (31.2.5)

for all E Borel. There also exists a Borel set of α measure zero N, such that for each x /∈N,there exists a Radon probability measure νx such that if f is a nonnegative µ measurablefunction or a µ measurable function in L1 (µ),

y→ f (x,y) is νx measurable α a.e.

x→∫Rm

f (x,y)dνx (y) is α measurable (31.2.6)

and ∫Rn+m

f (x,y)dµ =∫Rn

(∫Rm

f (x,y)dνx (y))

dα (x). (31.2.7)

If ν̂x is any other collection of Radon measures satisfying 31.2.6 and 31.2.7, then ν̂x = νxfor α a.e. x.

Proof:

Existence and uniqueness of α

First consider the uniqueness of α . Suppose α1 is another Radon measure satisfying31.2.5. Then in particular, α1 and α agree on open sets and so the two measures are thesame by Lemma 31.2.2.

To establish the existence of α , define α0 on Borel sets by

α0 (E) = µ (E×Rm).

Thus α0 is a finite Borel measure and so it is finite on compact sets. Lemma 14.2.3 on Page388 implies the existence of the Radon measure α extending α0.

Uniqueness of νx

Next consider the uniqueness of νx. Suppose νx and ν̂x satisfy all conclusions of thetheorem with exceptional sets denoted by N and N̂ respectively. Then, enlarging N and N̂,one may also assume, using Lemma 31.1.1, that for x /∈ N ∪ N̂, α (B(x,r)) > 0 wheneverr > 0. Now let

A =m

∏i=1

(ai,bi]