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Positive real powers preserve measurability

Statement

Let $r > 0$ . The map $t \mapsto t^{r}$ is strictly increasing on $[0, \infty)$ . Consequently, if $(X, A)$ is a measurable space and $h : X \to [0, \infty)$ is measurable, then $h^{r}$ is measurable.

Facts & Assumptions

Given: A measurable $h : X \to [0, \infty)$ and a real exponent $r > 0$ .

[F1]

For $r > 0$ , $t \mapsto t^{r}$ is continuous and differentiable on $(0, \infty)$ with derivative $r t^{r - 1}$ ; also $0^{r} = 0$ and $t^{r} > 0$ for $t > 0$ (Continuity and derivatives of positive-base real powers, Real powers for positive bases, with the zero-base positive-exponent convention).

[F2]

Positive-base real powers obey $(a^{s})^{r} = a^{s r}$ , so $(c^{1 / r})^{r} = c$ for $c > 0$ (The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents).

[F3]

A real-valued function is measurable exactly when all threshold sets of the form ${x : g (x) > c}$ are measurable (Threshold characterisations of real-valued and extended-real-valued measurability, A measurable function between measurable spaces).

[F4]

The mean value theorem applies on every compact interval where the function is continuous on the interval and differentiable in its interior (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

Given: A real exponent $r > 0$ and, for the measurability clause, a measurable function $h : X \to [0, \infty)$ .

1.1F1F4algebra

The map $t \mapsto t^{r}$ is strictly increasing on $[0, \infty)$ . If $0 < s < t$ , [F1] gives continuity on $[s, t]$ and positive derivative $r u^{r - 1}$ at every $u \in (s, t)$ , so [F4] gives $s^{r} < t^{r}$ ; if $s = 0 < t$ , [F1] gives $s^{r} = 0 < t^{r}$ .

2.1F2F3step 1.1given∎

For a threshold $c < 0$ , ${h^{r} > c} = X$ . For $c = 0$ , the set is ${h > 0}$ . For $c > 0$ , [F2] and strict increase from step 1.1 give ${h^{r} > c} = {h > c^{1 / r}}$ . Each such set is measurable because $h$ is measurable, so [F3] implies $h^{r}$ is measurable.

Depends on

Used by

Dependency tree · two levels

29 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

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)