1084 CHAPTER 31. DIFFERENTIATION, RADON MEASURES

31.2 Slicing MeasuresLet µ be a finite Radon measure. I will show here that a formula of the following formholds.

µ (F) =∫

Fdµ =

∫Rn

∫Rm

XF (x,y)dνx (y)dα (x)

where α (E) = µ (E×Rm). When this is done, the measures, νx, are called slicing mea-sures and this shows that an integral with respect to µ can be written as an iterated integralin terms of the measure α and the slicing measures, νx. This is like going backwards inthe construction of product measure. One starts with a measure µ , defined on the Cartesianproduct and produces α and an infinite family of slicing measures from it whereas in theconstruction of product measure, one starts with two measures and obtains a new measureon a σ algebra of subsets of the Cartesian product of two spaces. These slicing measuresare dependent on x. Later, this will be tied to the concept of independence or not of randomvariables. First here are two technical lemmas.

Lemma 31.2.1 The space Cc (Rm) with the norm

|| f || ≡ sup{| f (y)| : y ∈ Rm}

is separable.

Proof: Let Dl consist of all functions which are of the form

∑|α|≤N

aα yα

(dist(

y,B(0,l +1)C))nα

where aα ∈Q, α is a multi-index, and nα is a positive integer. Consider D ≡∪lDl . Then Dis countable. If f ∈Cc (Rn) , then choose l large enough that spt( f )⊆ B(0,l +1), a locallycompact space, f ∈ C0 (B(0, l +1)). Then since Dl separates the points of B(0,l +1) isclosed with respect to conjugates, and annihilates no point, it is dense in C0 (B(0, l +1)) bythe Stone Weierstrass theorem. Alternatively, D is dense in C0 (Rn) by Stone Weierstrassand Cc (Rn) is a subspace so it is also separable. So is Cc (Rn)+ , the nonnegative functionsin Cc (Rn).

From the regularity of Radon measures, the following lemma follows.

Lemma 31.2.2 If µ and ν are two Radon measures defined on σ algebras, Sµ and Sν , ofsubsets of Rn and if µ (V ) = ν (V ) for all V open, then µ = ν and Sµ = Sν .

Proof: Every compact set is a countable intersection of open sets so the two measuresagree on every compact set. Hence it is routine that the two measures agree on every Gδ andFσ set. (Recall Gδ sets are countable intersections of open sets and Fσ sets are countableunions of closed sets.) Now suppose E ∈ Sν is a bounded set. Then by regularity of ν

there exists G a Gδ set and F, an Fσ set such that F ⊆ E ⊆ G and ν (G\F) = 0. Then it isalso true that µ (G\F) = 0. Hence E = F ∪ (E \F) and E \F is a subset of G\F, a set ofµ measure zero. By completeness of µ, it follows E ∈Sµ and

µ (E) = µ (F) = ν (F) = ν (E) .