Proof of The Cell Integral of a Translated Periodic Function

Each result cited is universally quantified over the data in its own statement. The sum $y+h$ and the difference $y-x$ of points of $\mathbb{R}^{n}$ are formed coordinatewise, by Sum of Points of $\mathbb{R}^n$ and Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, and two points of $\mathbb{R}^{n}$ are equal exactly when their coordinates agree, by claim 1 of Euclidean Points as Tuples of Real Numbers; hence identities such as $(y+m)-x=(y-x)+m$, $(y-x)+x=y$ and $(z+x)-x=z$, and the cancellation $y+a=q+a\Rightarrow y=q$, hold in $\mathbb{R}^{n}$ because they hold coordinatewise in the field $\mathbb{R}$. For $A\subseteq\mathbb{R}^{n}$ and $h\in\mathbb{R}^{n}$ write $A+h=\{z+h:z\in A\}$, as in Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ; $\mathbf{1}_{A}v$ denotes the pointwise product of the indicator with a map $v$.

Periodicity of $w_{x}$. Let $y\in\mathbb{R}^{n}$ and $m\in\mathbb{Z}^{n}$. Then $(y+m)-x=(y-x)+m$, so $w_{x}(y+m)=w((y-x)+m)=w(y-x)=w_{x}(y)$ by the periodicity of $w$.

Measurability of $w_{x}$. Let $B$ be a Borel subset of $\mathbb{R}$. Then $w^{-1}(B)\in\mathcal{B}(\mathbb{R}^{n})$ since $w$ is measurable, and

$$w_{x}^{-1}(B)=\{y\in\mathbb{R}^{n}:y-x\in w^{-1}(B)\}=w^{-1}(B)+x,$$

because $y-x\in w^{-1}(B)$ holds exactly when $y=z+x$ for some $z\in w^{-1}(B)$, namely $z=y-x$. The set $w^{-1}(B)+x$ is Borel by claim 1 of Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ. Hence $w_{x}$ is measurable by Measurable Map, and the Set of Measurable Maps.

Integrability and the integral. Let $a=-x$ be the additive inverse of $x$, the point with coordinates $-x_{i}$ by claim 2 of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, so that $y+a=y-x$ for every $y\in\mathbb{R}^{n}$ by claim 3 of that proposition. By The Half-Open Unit Cell Tiles Euclidean Space (v), applied to $w$ and $h=a$, the map $\mathbf{1}_{Q+a}\,w$ is integrable and

$$\int_{\mathbb{R}^{n}}\mathbf{1}_{Q+a}\,w\,d\lambda_{n}=\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}\,w\,d\lambda_{n} .$$

Let $f=\mathbf{1}_{Q+a}\,w$, which is measurable, since Integrable Functions and Their Lebesgue Integral defines integrability only for measurable maps. For $y\in\mathbb{R}^{n}$ one has $f(y+a)=\mathbf{1}_{Q+a}(y+a)\,w(y+a)=\mathbf{1}_{Q}(y)\,w_{x}(y)$: indeed $y+a=y-x$, and $y+a\in Q+a$ holds exactly when $y+a=q+a$ for some $q\in Q$, that is, by cancellation, exactly when $y\in Q$. So the map $y\mapsto f(y+a)$ is $\mathbf{1}_{Q}\,w_{x}$. By claim 3 of Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ, applied to $f$ and $a$, the map $y\mapsto f(y+a)$ is integrable, since $f$ is, and

$$\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}\,w_{x}\,d\lambda_{n}=\int_{\mathbb{R}^{n}}f(y+a)\,d\lambda_{n}(y)=\int_{\mathbb{R}^{n}}f\,d\lambda_{n}=\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}\,w\,d\lambda_{n} .$$