I am trying to prove that lesbegue outer measure is translation invariant, i.e., $m^\ast (E+y)=m^\ast E$.
I proceed as follows. Let $E$ be a set. Let $\{I_n\}$ be a collection of open intervals that cover E. It follows that $\{I_n+y\}$ covers $E+y$. Then for any $\epsilon>0$, there is $\{I_n\}$ $$m^\ast E+\epsilon>\sum l(I_n)=\sum l(I_n+y)\geq m^\ast (E+y).$$ This implies that $$m^\ast E\geq m^\ast (E+y).$$ To finish the proof, I must get $$m^\ast E\leq m^\ast (E+y),$$ by making use of $$E=(E+y)-y.$$ I am failing to figure this out.
Best Answer
If $\ell (I)$ denotes the length of the interval $I$, remember that: $${\mathfrak{m}}^*E = \inf\left\{ \sum_{j \geq 1} \ell(I_j) \ : \ (I_j)_{j \geq 1} \mbox{ are open intervals and } E \subseteq \bigcup_{j \geq 1} I_j\right\}$$ If I call, for simplicity the above set $\mathscr{A}$, and $\mathscr{B}$ the same with respect to $E + y$, we can prove that $\mathscr{A = B}$.
Let $x \in \mathscr{A}$. So, we have $x = \sum_{j \geq 1} \ell (I_j)$ for some open intervals $I_j$ such that $E \subseteq \bigcup_{j \geq 1} I_j$. Now, that gives: $$E + y \subseteq \Big( \bigcup_{j \geq 1} I_j \Big) + y = \bigcup_{j \geq 1} (I_j +y)$$ Notice that the $I_j + y$ are also open, and $\ell(I_j) = \ell(I_j + y)$. This way we have: $$x = \sum_{j \geq 1} \ell(I_j) = \sum_{j \geq 1} \ell(I_j + y) \in \mathscr{B}$$ and hence $\mathscr{A} \subseteq \mathscr{B}$. The same argument, starting with $E + y$ instead of $E$ and $(-y)$ instead of $y$ gives $\mathscr{B \subset A}$. So we finally have $\mathscr{A = B}$, that is, $\inf \mathscr{A} = \inf \mathscr{B}$, which is exactly the equality $\mathfrak{m}^*E = \mathfrak{m}^*(E+y)$ that we wanted.