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Threshold characterisations of real-valued and extended-real-valued measurability

Statement

Let $(X, A)$ be a measurable space and let $f : X \to \overline{R}$ . The following are equivalent:

$f$ is measurable;
${x : f (x) > a} \in A$ for every real $a$ ;
${x : f (x) \geq a} \in A$ for every real $a$ ;
${x : f (x) < a} \in A$ for every real $a$ ;
${x : f (x) \leq a} \in A$ for every real $a$ .

Moreover, in any one of conditions 2 through 5 it is enough to test only rational thresholds $a \in Q$ .

Facts & Assumptions

Given: A measurable space $(X, A)$ and a function $f : X \to \overline{R}$ .

[L1]

The Borel sigma-algebra on $\overline{R}$ is generated by the rays $(a, + \infty]$ with $a \in R$ . (The Borel sigma-algebra on the extended real line)

[L2]

A generating family on the codomain suffices to test measurability. (A generating family on the codomain suffices to test measurability)

[L3]

Between any two distinct real numbers there lies a rational number. (The rationals embed densely in the reals)

Proof

technique · direct

1.1

By [L1], the threshold set ${x : f (x) > a}$ is the preimage [L1, L2] $f^{- 1} ((a, + \infty])$ . Therefore [L2] gives the equivalence of condition 1 and condition 2.

L1L2

1.2

Suppose the sets ${f > q}$ are measurable for every rational $q$ . For a [L3, algebra] real $a$ ,

${f \leq a} = X ∖ {f > a}, {f < a} = X ∖ {f \geq a} .$

So conditions 2 and 5 are equivalent, and conditions 3 and 4 are equivalent. [step 1.1, algebra]

2.1

For every real $a$ ,

step 1.1algebra

${f \geq a} = ⋂_{n = 1}^{\infty} {f > a - 1 / n}, {f > a} = ⋃_{n = 1}^{\infty} {f \geq a + 1 / n},$

so conditions 2 and 3 are equivalent. Combining this with steps 1.1 and 1.2 shows that conditions 1 through 5 are all equivalent. [step 1.1, step 1.2, algebra]

2.2

For every real $a$ ,

step 1.1algebra

${f > a} = ⋃_{q \in Q, q > a} {f > q} .$

Indeed, $f (x) > q > a$ implies $f (x) > a$ , and if $f (x) > a$ then [L3] gives a rational $q$ with $a < q < f (x)$ unless $f (x) = + \infty$ , in which case any rational $q > a$ works. Thus the real-threshold version of condition 2 follows from the rational one. The converse is immediate, so in condition 2 it is enough to test only rational thresholds. [L3, algebra]

3.1

The equivalences from steps 2.1 and 2.2 transfer the rational-threshold [step 2.1, step 2.2, step 1.2] reduction of step 1.2 to conditions 3 through 5. Therefore in any one of conditions 2 through 5 it is enough to test only rational thresholds.

step 2.1step 2.2step 1.2∎

Depends on

Used by

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Measure, Integration and Real Analysis, Proposition 2.52 (standard reference, not scraped)
John K. Hunter, Measure Theory, Propositions 3.4 and 3.5 (standard reference, not scraped)