Exercise 1.2.20 (Translation invariance of the Lebesgue measure)

The cornerstone of this proof is to use a similar result about elementary sets proven in Exercise 1.1.6.

(i)

$E+x$is Lebesgue measurable
We demonstrate that for any $𝜖>0$ there exists an open set $U^{\prime }$ such that $m^{\text{∗}}(U^{\prime }∕(E+x))\le 𝜖$. Fix $𝜖>0$ and take an open set $U$ such that $m(U∕E)\le 𝜖$. Notice that (1) the set $U+x$ covers $E+x$ and that (2) the set $U∕E+x=(U+x)∕(E+x)$. Thus $U^{\prime }=U+x$ is a good candidate. Let $B$ be an elementary set such that $m(B△[U∕E])\le 𝜖$ by Exercise 1.2.6. This implies that $(B+x)△[(U+x)∕(E+x)]\le 𝜖$. By Exercise 1.2.6. $(U+x)∕(E+x)$ is a Lebesgue-measurable, and we have $m((U+x)∕(E+x))\le 𝜖$.

(ii)

$m(E+x)=m(E)$
We want to demonstrate that $$m(E+x)=\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\underset{n=1}{\overset{\infty }{\sum }}|B_n|:\begin{array}{c}{B}_n\text{ are boxes}\hfill \\ E+x\subset \underset{n=1}{\overset{\infty }{\bigcup }}{B}_n\hfill \end{array}\right\}=\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\underset{n=1}{\overset{\infty }{\sum }}|B_n|:\begin{array}{c}{B}_n\text{ are boxes}\hfill \\ E\subset \underset{n=1}{\overset{\infty }{\bigcup }}{B}_n\hfill \end{array}\right\}=m(E)$$

We do it by demonstrating

$$\left\{\underset{n=1}{\overset{\infty }{\sum }}|B_n|:\begin{array}{c}{B}_n\text{ are boxes}\hfill \\ E+x\subset \underset{n=1}{\overset{\infty }{\bigcup }}{B}_n\hfill \end{array}\right\}=\left\{\underset{n=1}{\overset{\infty }{\sum }}|B_n|:\begin{array}{c}{B}_n\text{ are boxes}\hfill \\ E\subset \underset{n=1}{\overset{\infty }{\bigcup }}{B}_n\hfill \end{array}\right\}$$

But this is evident from the fact that we can get a collection of boxes covering $E+x$ by taking a collection of boxes covering $E$ and adding $x$ to each box; similarly taking cover of $E+x$ we can turn it into the cover of $E$ by subtracting an $x$ from it. The equality of the measures then follows by the translation invariance of the elementary measures.