Proof of Translation Invariance of Lebesgue Measure and the Lebesgue Integral

Fix a real number $t$ and write $\tau_t:\mathbb{R}\to\mathbb{R}$, $\tau_t(x)=x+t$, so that $\tau_t\circ\tau_{-t}$ and $\tau_{-t}\circ\tau_t$ are the identity and $A+t=\tau_{-t}^{-1}(A)$ for every $A\subseteq\mathbb{R}$.

Claim 1. Let $A\subseteq\mathbb{R}$ and let $\bigl((a_m,b_m)\bigr)_{m\in\mathbb{N}}$ be a sequence of open intervals covering $A$ as in Lebesgue Outer Measure on the Real Line. Then the intervals $(a_m+t,\,b_m+t)$ cover $A+t$: if $x\in A+t$ then $x-t\in A$, so $a_m<x-t<b_m$ for some $m$, i.e., $a_m+t<x<b_m+t$; and each has the same length $(b_m+t)-(a_m+t)=b_m-a_m$. Thus every covering of $A$ induces a covering of $A+t$ with the same total length, so $\lambda^{*}(A+t)\le\lambda^{*}(A)$; applying this inequality with $A+t$ in place of $A$ and $-t$ in place of $t$, and using $(A+t)+(-t)=A$, gives the reverse inequality. Hence $\lambda^{*}(A+t)=\lambda^{*}(A)$.

Next, the family $\mathcal{D}=\{B\subseteq\mathbb{R}:\ B+t\in\mathcal{B}(\mathbb{R})\}$ is a $\sigma$-algebra: $\mathbb{R}+t=\mathbb{R}$, $(\mathbb{R}\setminus B)+t=\mathbb{R}\setminus(B+t)$, and $\bigl(\bigcup_m B_m\bigr)+t=\bigcup_m(B_m+t)$. It contains every open subset $U$ of $\mathbb{R}$: if $x\in U+t$ then $x-t\in U$, so there is $\varepsilon>0$ with $(x-t-\varepsilon,\,x-t+\varepsilon)\subseteq U$, and then $(x-\varepsilon,\,x+\varepsilon)=(x-t-\varepsilon,\,x-t+\varepsilon)+t\subseteq U+t$, so $U+t$ is open, hence Borel. Since $\mathcal{B}(\mathbb{R})$ is the $\sigma$-algebra generated by the open sets, $\mathcal{B}(\mathbb{R})\subseteq\mathcal{D}$; that is, $B+t$ is Borel whenever $B$ is. Since $\lambda$ is the restriction of $\lambda^{*}$ to $\mathcal{B}(\mathbb{R})$ by Existence of Lebesgue Measure on the Real Line, $\lambda(B+t)=\lambda^{*}(B+t)=\lambda^{*}(B)=\lambda(B)$.

Claim 2. The function $f\circ\tau_t$ (that is, $x\mapsto f(x+t)$) is measurable: for Borel $B'$, $(f\circ\tau_t)^{-1}(B')=\tau_t^{-1}\bigl(f^{-1}(B')\bigr)=f^{-1}(B')+(-t)$, which is Borel by Claim 1. For a nonnegative simple function $s=\sum_i c_i\mathbf{1}_{A_i}$ with Borel $A_i$ and $c_i\ge0$, the composition $s\circ\tau_t=\sum_i c_i\mathbf{1}_{A_i+(-t)}$ is again a nonnegative simple function, and its integral is $\sum_i c_i\,\lambda\bigl(A_i+(-t)\bigr)=\sum_i c_i\,\lambda(A_i)$ by Claim 1; that is, simple functions and their translates have equal integrals. Moreover $s\mapsto s\circ\tau_t$ is a bijection from the set of simple functions $s'$ with $0\le s'\le f$ pointwise onto the set of simple functions $s$ with $0\le s\le f\circ\tau_t$ pointwise, with inverse $s\mapsto s\circ\tau_{-t}$: the compositions are inverse to one another because $\tau_t\circ\tau_{-t}$ is the identity, and $0\le s'\le f$ holds pointwise if and only if $0\le s'\circ\tau_t\le f\circ\tau_t$ does (evaluate at $x+t$ as $x$ ranges over $\mathbb{R}$). Hence the two suprema defining the integrals of $f\circ\tau_t$ and of $f$ range over equal sets of real numbers, and