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For sigma-finite measures, the section-measure functions are measurable

Statement

Let $(X, A, μ)$ and $(Y, B, ν)$ be sigma-finite measure spaces, and let $E \in A \otimes B$ . Then the functions

$x \mapsto ν (E_{x}), y \mapsto μ (E^{y})$

are measurable from $X$ and $Y$ into $[0, \infty]$ .

Facts & Assumptions

Given: Sigma-finite measure spaces $(X, A, μ)$ and $(Y, B, ν)$ , and a set $E \in A \otimes B$ .

[L1]

Finite disjoint unions of measurable rectangles form an algebra that generates $A \otimes B$ . (Finite disjoint unions of measurable rectangles form an algebra generating the product sigma-algebra)

[L2]

If an algebra generates a sigma-algebra, then its monotone class is that same sigma-algebra. (The monotone class generated by an algebra equals the sigma-algebra it generates)

[L3]

Pointwise monotone limits of measurable functions are measurable. (Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable)

[A1]

Since $ν$ is sigma-finite, there are measurable sets $Y_{m} ↑ Y$ with $ν (Y_{m}) < \infty$ for every $m$ . Likewise there are measurable $X_{n} ↑ X$ with $μ (X_{n}) < \infty$ .

[A2]

If $F_{k} ↑ F$ , then $(F_{k})_{x} ↑ F_{x}$ and $ν ((F_{k})_{x} \cap Y_{m}) ↑ ν (F_{x} \cap Y_{m})$ for every $x$ . If $F_{k} ↓ F$ , then $(F_{k})_{x} \cap Y_{m} ↓ F_{x} \cap Y_{m}$ , and continuity from above on the finite-measure space $Y_{m}$ gives $ν ((F_{k})_{x} \cap Y_{m}) ↓ ν (F_{x} \cap Y_{m})$ .

Proof

technique · direct

1.1

Fix $m \geq 1$ and let $C_{m}$ be the family of sets $F \in A \otimes B$ for which $x \mapsto ν (F_{x} \cap Y_{m})$ is $A$ -measurable. If $F = A \times B$ is a measurable rectangle, then $ν (F_{x} \cap Y_{m}) = ν (B \cap Y_{m}) 1_{A} (x),$ so $F \in C_{m}$ . By [A2], $C_{m}$ is a monotone class. Hence [L1] and [L2] imply that every product-measurable set lies in $C_{m}$ .

L1L2A1A2

2.1

Applying step 1.1 to the given set $E$ shows that $g_{m} (x) : = ν (E_{x} \cap Y_{m})$ is measurable for every $m$ . Because $Y_{m} ↑ Y$ , one has $g_{m} (x) ↑ ν (E_{x})$ for each $x$ , so [L3] gives measurability of $x \mapsto ν (E_{x})$ .

A1L3step 1.1

3.1

The same argument with the finite-measure exhaustion $X_{n} ↑ X$ shows that $y \mapsto μ (E^{y})$ is measurable. Therefore both section-measure functions are measurable.

A1L3step 1.1step 2.1∎

Depends on

Used by

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Sources

Terence Tao, An Introduction to Measure Theory, Corollary 1.7.17 (standard reference, not scraped)
John K. Hunter, Measure Theory, Theorem 5.15 (standard reference, not scraped)