Proof of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$

Points of $\mathbb{R}^n$ are written in coordinates, so $a=(a_1,\dots,a_n)$, $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$. Let $T(x)=x+a$ and $S(x)=a-x$, and let $d$ be the Euclidean distance. The index $j$ always ranges over $\{1,\dots,n\}$. Throughout, $\lambda$ is Lebesgue measure on the real line and $B-a$ abbreviates $B+(-a)$.

Step 1: $T$ and $S$ are Borel measurable. The difference $x-y$ has $j$-th coordinate $x_j-y_j$, its Euclidean norm is $d(x,y)$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, and claim 4 of the same lemma bounds each coordinate of a point by its norm; together these give $|x_j-y_j|\le d(x,y)$ for every $j$. Hence, if $(x^k)_{k\in\mathbb{N}}$ is a sequence in $\mathbb{R}^n$ and $x\in\mathbb{R}^n$ with $d(x^k,x)\to0$, then $x_j^k\to x_j$ for every $j$, and therefore $x_j^k+a_j\to x_j+a_j$ and $a_j-x_j^k\to a_j-x_j$. So every component of $T$ and of $S$ is sequentially continuous in the sense of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, and claim 3(b) there makes $T$ and $S$ measurable with respect to the $\sigma$-algebra $\mathcal{B}_n$ of that lemma, which is $\mathcal{B}(\mathbb{R}^n)$ by claim 5 there.

Step 2: preimages. For any $B\subseteq\mathbb{R}^n$ one has $T^{-1}(B)=B-a$ and $S^{-1}(B)=a-B$: indeed $x+a\in B$ holds exactly when $x=(x+a)-a$ lies in $B-a$, and $a-x\in B$ holds exactly when $x=a-(a-x)$ lies in $a-B$. Consequently, for Borel $B$ the set $a-B=S^{-1}(B)$ is Borel by step 1, and so is $B+a$, being the preimage of $B$ under the map $x\mapsto x-a$, which is measurable by step 1 applied with $-a$ in place of $a$.

Step 3: invariance on sets. Let $\mu$ be the image measure of $\lambda_n$ under $T$, a measure on $\mathcal{B}(\mathbb{R}^n)$ by claim 1 of that lemma; by the definition of the image measure and step 2, $\mu(B)=\lambda_n\bigl(T^{-1}(B)\bigr)=\lambda_n(B-a)$ for every Borel $B$. Let $A_1,\dots,A_n$ be Borel subsets of $\mathbb{R}$. A point $x$ satisfies $x+a\in A_1\times\dots\times A_n$ exactly when $x_j\in A_j-a_j$ for every $j$, so

each factor being Borel with $\lambda(A_j-a_j)=\lambda(A_j)$ by Translation Invariance of Lebesgue Measure and the Lebesgue Integral. Hence

These are exactly the values that characterize $\lambda_n$ among measures on $\mathcal{B}(\mathbb{R}^n)$ in Lebesgue Measure on $\mathbb{R}^n$, and only one measure has them; therefore $\mu=\lambda_n$, that is, $\lambda_n(B-a)=\lambda_n(B)$ for every Borel $B$. Applying this with $-a$ in place of $a$ gives $\lambda_n(B+a)=\lambda_n(B)$.

Let $\nu$ be the image measure of $\lambda_n$ under $S$. In the same way $S^{-1}(A_1\times\dots\times A_n)=(a_1-A_1)\times\dots\times(a_n-A_n)$, and $a_j-A_j=(-A_j)+a_j$, so $\lambda(a_j-A_j)=\lambda(-A_j)=\lambda(A_j)$, the first equality by Translation Invariance of Lebesgue Measure and the Lebesgue Integral and the second by claim 1 of Reflection Invariance of Lebesgue Measure and Symmetry of the Standard Normal Distribution, which also gives that $-A_j$ is Borel. As before $\nu=\lambda_n$, that is, $\lambda_n(a-B)=\lambda_n(B)$ for every Borel $B$. This proves claim 1.

Step 4: integrals. Let $f:\mathbb{R}^n\to[0,\infty]$ be measurable, which by Lebesgue Integral of a Nonnegative Measurable Function means that $\{y\in\mathbb{R}^n:f(y)>c\}$ is Borel for every real number $c$. For every real $c$,

which is Borel by step 1; hence $f\circ T$ is measurable, and the same argument with $S$ in place of $T$ shows that $f\circ S$ is measurable. By step 3 the image measure of $\lambda_n$ under $T$ is $\lambda_n$ itself, so claim 2 of Image Measures, Measures with Densities, and Change of Variables gives

and the same argument with $S$ in place of $T$ gives the identity for $x\mapsto f(a-x)$. This proves claim 2.

For claim 3, let $f:\mathbb{R}^n\to\mathbb{R}$ be measurable. The set $\{y\in\mathbb{R}^n:f(y)>c\}$ is then Borel for every real $c$, being the preimage under $f$ of a ray, and every ray belongs to $\mathcal{B}(\mathbb{R})$ because by claim 2 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line the rays generate $\mathcal{B}(\mathbb{R})$ and are therefore contained in it; so the displayed identity above again shows that $f\circ T$ and $f\circ S$ are measurable. The second half of claim 2 of Image Measures, Measures with Densities, and Change of Variables, applied to $T$ and to $S$ exactly as above, states that a measurable real-valued $f$ is integrable with respect to the image measure, here $\lambda_n$, if and only if $f\circ T$, respectively $f\circ S$, is integrable with respect to $\lambda_n$, and that in that case the displayed identities hold in $\mathbb{R}$.