Proof of Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ

For $b\in\mathbb{R}^{n}$ let $T_{b}:\mathbb{R}^{n}\to\mathbb{R}^{n}$, $T_{b}(x)=x+b$, and let $S:\mathbb{R}^{n}\to\mathbb{R}^{n}$, $S(x)=-x$; componentwise, $T_{b}(x)_{i}=x_{i}+b_{i}$ and $S(x)_{i}=-x_{i}$.

Step 1: boxes and measurability. For $c,d\in\mathbb{R}^{n}$,

$$T_{b}^{-1}\bigl(R(c,d)\bigr)=R(c-b,\,d-b),\qquad S^{-1}\bigl(R(c,d)\bigr)=R(-d,\,-c),$$

since $c_{i}<x_{i}+b_{i}<d_{i}$ if and only if $c_{i}-b_{i}<x_{i}<d_{i}-b_{i}$, and $c_{i}<-x_{i}<d_{i}$ if and only if $-d_{i}<x_{i}<-c_{i}$. Open boxes lie in $\mathcal{B}(\mathbb{R}^{n})$ and generate it (Existence and Uniqueness of Lebesgue Measure on ℝⁿ (i)), so $T_{b}$ and $S$ are $\mathcal{B}(\mathbb{R}^{n})/\mathcal{B}(\mathbb{R}^{n})$-measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (i)).

Step 2: invariance of $\lambda_{n}$. The image measures $\lambda_{n}\circ T_{b}^{-1}$ and $\lambda_{n}\circ S^{-1}$ are measures on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (i)). Let $c,d\in\mathbb{R}^{n}$ with $c_{i}<d_{i}$ for every $i\in[n]$; then also $c_{i}-b_{i}<d_{i}-b_{i}$ and $-d_{i}<-c_{i}$, and by Step 1 and Lebesgue Measure on $\mathbb{R}^n$ (i),

$$(\lambda_{n}\circ T_{b}^{-1})\bigl(R(c,d)\bigr)=\prod_{i=1}^{n}\bigl((d_{i}-b_{i})-(c_{i}-b_{i})\bigr)=\prod_{i=1}^{n}(d_{i}-c_{i}),$$ $$(\lambda_{n}\circ S^{-1})\bigl(R(c,d)\bigr)=\prod_{i=1}^{n}\bigl((-c_{i})-(-d_{i})\bigr)=\prod_{i=1}^{n}(d_{i}-c_{i}).$$

These are the values that characterise $\lambda_{n}$, so by the uniqueness in Existence and Uniqueness of Lebesgue Measure on ℝⁿ (iii),

$$\lambda_{n}\circ T_{b}^{-1}=\lambda_{n}\quad\text{for every }b\in\mathbb{R}^{n},\qquad\lambda_{n}\circ S^{-1}=\lambda_{n}.\tag{$*$}$$

Step 3: sets. For $B\subseteq\mathbb{R}^{n}$, $x\in T_{-a}^{-1}(B)$ means $x-a\in B$, that is, $x=y+a$ with $y\in B$; and $x\in S^{-1}(B)$ means $-x\in B$, that is, $x\in-B$. So

$$B+a=T_{-a}^{-1}(B),\qquad -B=S^{-1}(B).$$

For $B\in\mathcal{B}(\mathbb{R}^{n})$ both sets are therefore Borel by Step 1, and by $(*)$

$$\lambda_{n}(B+a)=(\lambda_{n}\circ T_{-a}^{-1})(B)=\lambda_{n}(B),\qquad\lambda_{n}(-B)=(\lambda_{n}\circ S^{-1})(B)=\lambda_{n}(B).$$

This proves (i) and (ii).

Step 4: functions. The functions $x\mapsto f(x+a)$ and $x\mapsto f(-x)$ are $f\circ T_{a}$ and $f\circ S$. For $f\in\mathcal{M}^{+}(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (ii) shows that both lie in $\mathcal{M}^{+}(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, and together with $(*)$ it gives

$$\int_{\mathbb{R}^{n}}f\circ T_{a}\,d\lambda_{n}=\int_{\mathbb{R}^{n}}f\,d(\lambda_{n}\circ T_{a}^{-1})=\int_{\mathbb{R}^{n}}f\,d\lambda_{n},$$

and likewise with $S$ in place of $T_{a}$; this proves (iii). For measurable $f:\mathbb{R}^{n}\to\mathbb{R}$, Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (iii) and $(*)$ show that $f\circ T_{a}$ is measurable and that $f\circ T_{a}\in\mathcal{L}^{1}(\lambda_{n})$ if and only if $f\in\mathcal{L}^{1}(\lambda_{n}\circ T_{a}^{-1})=\mathcal{L}^{1}(\lambda_{n})$, with

$$\int_{\mathbb{R}^{n}}f\circ T_{a}\,d\lambda_{n}=\int_{\mathbb{R}^{n}}f\,d\lambda_{n}$$

in that case, and likewise with $S$. This proves (iv).