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For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique

Statement

Let (X,A,μ) and (Y,B,ν) be sigma-finite measure spaces. Then:

  1. the set function E↦(μ×ν)(E) of The product measure of two sigma-finite measure spaces is a measure on A⊗B;
  2. for measurable rectangles, (μ×ν)(A×B)=μ(A)ν(B), with the nonnegative extended-product convention 0⋅∞=∞⋅0=0;
  3. the measure μ×ν is sigma-finite; and
  4. it is the unique measure on A⊗B with the rectangle formula.

Facts & Assumptions

Given: Sigma-finite measure spaces (X,A,μ) and (Y,B,ν).

[L1]

For every product-measurable set E, (μ×ν)(E)=∫Xν(Ex) dμ=∫Yμ(Ey) dν. (The product measure of two sigma-finite measure spaces)

[L2]

Monotone convergence passes increasing limits through nonnegative integrals. (Monotone convergence for the integral)

[L3]

A measure is determined by its values on a sigma-finite generating pi-system. (Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system)

[L4]

The nonnegative integral is additive for finite sums of nonnegative measurable functions. (Additivity of the nonnegative Lebesgue integral)

[L5]

The nonnegative integral agrees with the simple integral, whose zero coefficient on a null cell contributes zero even when another factor is infinite (The nonnegative integral agrees with the simple integral on simple functions, The integral of a nonnegative simple function).

[A1]

Measurable rectangles form a pi-system that generates A⊗B.

[A2]

Since μ and ν are sigma-finite, there are measurable exhaustions Xn↑X and Yn↑Y with μ(Xn),ν(Yn)<∞.

Proof

technique · direct
1.1L1L5

If E=A×B is a measurable rectangle, then (A×B)x={B,x∈A,∅,x∉A. Thus its section-measure function is ν(B)1A, where the value outside A is zero even when ν(B)=∞. The simple-function integral [L5] gives (μ×ν)(A×B)=∫Xν((A×B)x) dμ=μ(A)ν(B) with 0⋅∞=0 when μ(A)=0, and symmetrically when ν(B)=0. This proves the rectangle formula in every extended-value case.

1.2L1L2L4

Let E1,E2,… be pairwise disjoint measurable subsets of X×Y, and put FN:=⋃k≤NEk. Then (FN)x=⋃k≤N(Ek)x is a disjoint union, so ν((FN)x)=∑k≤Nν((Ek)x) for every x. The section functions are nonnegative measurable, and finite additivity of their integrals gives (μ×ν)(FN)=∫X∑k≤Nν((Ek)x) dμ=∑k≤N(μ×ν)(Ek). Since the section functions for FN increase pointwise to that of ⋃k≥1Ek, [L2] gives (μ×ν)(⋃k≥1Ek)=∑k≥1(μ×ν)(Ek). The empty set has product measure zero, so μ×ν is a measure.

2.1A2step 1.1

By [A2] and step 1.1, each rectangle Xn×Yn has finite product measure (μ×ν)(Xn×Yn)=μ(Xn)ν(Yn)<∞, and ⋃n(Xn×Yn)=X×Y. Hence μ×ν is sigma-finite.

3.1A1L3step 1.1step 2.1∎

Let ρ be another measure on A⊗B with the same rectangle formula. Step 1.1 shows that ρ and μ×ν agree on the generating pi-system of measurable rectangles, and step 2.1 gives the required sigma-finite exhaustion. Therefore [L3] implies ρ=μ×ν on all of A⊗B. This proves existence, the rectangle formula, sigma-finiteness, and uniqueness.

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