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

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here. The argument for claim 1 is the one given as Claim 1 in the published proof of The Integral over the Unit Cell of a Product of One-Variable Functions, recorded here as a standing item so that later results may cite it; the citation attached to this proof records that provenance.

Proof of claim 1. Measurability of a real-valued function is that of Measurable Map, and the Set of Measurable Maps: the preimage of every member of the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ lies in the $\sigma$-algebra of the domain. Let $B\in\mathcal{B}(\mathbb{R})$. If $0\notin B$ then $\tilde{f}^{-1}(B)=f^{-1}(B)$, and if $0\in B$ then $\tilde{f}^{-1}(B)=f^{-1}(B)\cup(X\setminus X_{0})$, since $\tilde{f}$ takes the value $0$ at every point outside $X_{0}$ and agrees with $f$ on $X_{0}$. By claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions every member of $\mathcal{F}|_{X_{0}}$ lies in $\mathcal{F}$, and $X\setminus X_{0}\in\mathcal{F}$ because $\mathcal{F}$ is a $\sigma$-algebra and $X_{0}\in\mathcal{F}$; so measurability of $f$ gives measurability of $\tilde{f}$ in both cases. Conversely $f^{-1}(B)=\tilde{f}^{-1}(B)\cap X_{0}$, which lies in $\mathcal{F}$ and is contained in $X_{0}$, hence lies in $\mathcal{F}|_{X_{0}}$; so measurability of $\tilde{f}$ gives measurability of $f$.

Assume now that both are measurable. Let $f^{+}$ and $f^{-}$ be the positive and negative parts of $f$ as in Integrable Functions and Their Lebesgue Integral, and $(\tilde{f})^{+}$ and $(\tilde{f})^{-}$ those of $\tilde{f}$. Since $\max\{0,0\}=0$, the map $(\tilde{f})^{+}$ is the zero extension of $f^{+}$ and $(\tilde{f})^{-}$ is the zero extension of $f^{-}$. All four maps are nonnegative, real-valued and measurable, the measurability by Integrable Functions and Their Lebesgue Integral; and for a nonnegative real-valued function, measurability in the sense of Measurable Map, and the Set of Measurable Maps and measurability as a map into $[0,\infty]$ agree, as recorded in Nonnegative Measurable Maps and Their Lebesgue Integral. Claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, applied to $f^{+}$ and then to $f^{-}$, therefore gives

$$\int_{X_{0}}f^{+}\,d\mu|_{X_{0}}=\int_{X}(\tilde{f})^{+}\,d\mu,\qquad \int_{X_{0}}f^{-}\,d\mu|_{X_{0}}=\int_{X}(\tilde{f})^{-}\,d\mu .$$

By Integrable Functions and Their Lebesgue Integral, $f$ is integrable with respect to $\mu|_{X_{0}}$ exactly when the two left-hand sides are finite, $\tilde{f}$ is integrable with respect to $\mu$ exactly when the two right-hand sides are finite, and in that case each of the two integrals in the claim is the difference of the corresponding pair of values. The two differences agree by the two displayed identities.

Proof of claim 2. By claim 1 of Restriction Stability of Continuity and of the Derivative, applied with $B=[0,1]\subseteq A=\mathbb{R}$, the restriction $h|_{[0,1]}$ is continuous on $[0,1]$. Let $\tilde{h}:\mathbb{R}\to\mathbb{R}$ be the map equal to $h$ on $[0,1]$ and to $0$ at every point outside $[0,1]$, that is, the zero extension of $h|_{[0,1]}$. By Agreement of the Riemann and Lebesgue Integrals for Continuous Functions on a Closed Interval, applied to the continuous map $h|_{[0,1]}$ on the closed interval determined by $0<1$, the map $h|_{[0,1]}$ is Riemann integrable on $[0,1]$ (claim 1 there), the map $\tilde{h}$ is measurable with respect to $\mathcal{B}(\mathbb{R})$ and integrable with respect to Lebesgue measure $\lambda$ (claim 2 there), and

$$\int_{\mathbb{R}}\tilde{h}\,d\lambda=\int_{0}^{1}h(t)\,dt$$

(claim 3 there).

Let $g:\mathbb{R}\to\mathbb{R}$ be the zero extension of $h|_{J}$, that is, the map equal to $h$ on $J$ and to $0$ at every point outside $J$. Since $J\subseteq[0,1]$, we have $g=\tilde{h}\,\mathbf{1}_{J}$, where $\mathbf{1}_{J}$ is the indicator of $J$: at a point of $J$ both sides equal $h(t)$, and at a point outside $J$ both sides are $0$. The set $J$ belongs to $\mathcal{B}(\mathbb{R})$ by The Integral over the Unit Cell of a Product of One-Variable Functions, so $\mathbf{1}_{J}$ is measurable by claim 1 of Arithmetic of Measurable Real- and [0,∞]-Valued Maps and $g$ is measurable by claim 3 of that lemma. By claim 1 of the present statement, applied to the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, to $X_{0}=J$ and to $f=h|_{J}$, whose zero extension is $g$, the map $h|_{J}$ is measurable with respect to $\mathcal{B}_{J}$; here the restriction of $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ to $J$ is $(J,\mathcal{B}_{J},\lambda_{J})$ by The Integral over the Unit Cell of a Product of One-Variable Functions.

The maps $g$ and $\tilde{h}$ agree at every point of $\mathbb{R}$ except possibly at $1$: they agree on $J$ and outside $[0,1]$ by their descriptions, and $[0,1]$ is the union of $J$ and $\{1\}$, since a real $t$ satisfies $0\le t\le1$ exactly when either $0\le t<1$ or $t=1$. The set $\{1\}$ is the closed interval determined by $1$ and $1$, so $\{1\}\in\mathcal{B}(\mathbb{R})$ by Borel Sigma-Algebra on the Real Line and $\lambda(\{1\})=0$ by claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness. Hence $g=\tilde{h}$ almost everywhere with respect to $\lambda$. Both maps are measurable and $\tilde{h}$ is integrable, so The Lebesgue Integral and Null Sets (iii) gives that $g$ is integrable with respect to $\lambda$ and that $\int_{\mathbb{R}}g\,d\lambda=\int_{\mathbb{R}}\tilde{h}\,d\lambda$.

By claim 1 of the present statement again, $h|_{J}$ is integrable with respect to $\lambda_{J}$ and

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

which is the asserted identity.