Zero Extension of a Real-Valued Function, and the Unit-Cell Integral of a Continuous Function

Measurability of a real-valued function is that of Measurable Map, and the Set of Measurable Maps, and integrability and the integral of a real-valued function are those of Integrable Functions and Their Lebesgue Integral. Then the following hold.

1. (Zero extension)(i) Let $(X,\mathcal{F},\mu)$ be a measure space, let $X_{0}\in\mathcal{F}$, and let $(X_{0},\mathcal{F}|_{X_{0}},\mu|_{X_{0}})$ be the restriction of $(X,\mathcal{F},\mu)$ to $X_{0}$ given by claim 1 of that lemma. Let $f:X_{0}\to\mathbb{R}$ and let $\tilde{f}:X\to\mathbb{R}$ be the map equal to $f$ on $X_{0}$ and to $0$ at every point of $X$ outside $X_{0}$. Then $f$ is measurable with respect to $\mathcal{F}|_{X_{0}}$ if and only if $\tilde{f}$ is measurable with respect to $\mathcal{F}$. In that case $f$ is integrable with respect to $\mu|_{X_{0}}$ if and only if $\tilde{f}$ is integrable with respect to $\mu$, and then

$$\int_{X_{0}}f\,d\mu|_{X_{0}}=\int_{X}\tilde{f}\,d\mu .$$

2. (Continuous integrands over the half-open unit interval)(ii) Let $J=\{t\in\mathbb{R}:0\le t<1\}$ and let $(J,\mathcal{B}_{J},\lambda_{J})$ be as in The Integral over the Unit Cell of a Product of One-Variable Functions, let $[0,1]$ be the closed interval determined by $0$ and $1$, and let $h:\mathbb{R}\to\mathbb{R}$ be continuous on $\mathbb{R}$, the domain and the codomain carrying the absolute value metric. Writing $h|_{J}$ and $h|_{[0,1]}$ for the restrictions of $h$, the map $h|_{J}$ is measurable with respect to $\mathcal{B}_{J}$ and integrable with respect to $\lambda_{J}$, the map $h|_{[0,1]}$ is Riemann integrable on $[0,1]$, and

$$\int_{J}h|_{J}\,d\lambda_{J}=\int_{0}^{1}h(t)\,dt ,$$

the right-hand side being the Riemann integral of $h|_{[0,1]}$.