The Outer Lebesgue Measure is Translation Invariant and scales nicely with Dilation.

Define the sets $A+t = \{ x+t: x\in A \}$ and $\delta A = \{\delta x: x\in A\}$ . Let $m^*$ be, of course, the outer Lebesgue measure. Suppose that we have a set $A \subset \mathbb{R}$ . In this case, it is possible to cover $A$ with a sequence $(A_n)_{n=1}^\infty$ of elementary sets, that is $A \subset \bigcup A_n$ . Note that since $A_n$ is elementary, they may be written as a finite union of disjoint intervals $I_k$ , $A_n= \bigcup I_k$ , for every integer $n$ .

By definition, $m(I_k) = (b_k - a_k)$ , where $a_k, b_k$ are the end points of the interval $I_k$ . Now, clearly $A+t \subset \bigcup (A_n + t) \iff A \subset \bigcup A_n$ .

This implies that $A_n +t = \bigcup (I_k + t)$ . Note that $m(I_k + t) = (b_k +t - (a_k+t) = (b_k- a_k)$ . Hence, $m(A_n + t) = \sum_k^N m(I_k + t) = \sum_k^N m(I_k) = m(A_n)$ , but the additivity of $m$ .

Fix $\epsilon > 0$ . Now, it is possible to choose these elementary coverings of $A$ in such a way (by definition of $m^*$ ) so that

$m^*(A+t) \leq \sum_n^\infty m(A_n + t) \leq m^*(A+t) + \epsilon$ .

But also

$m^*(A) \leq \sum_n^\infty m(A_n) \leq m^*(A) + \epsilon$ .

But, it was shown that $m(A_n) = m(A_n + t)$ .

Hence,

$m^*(A) \leq \sum_n^\infty m(A_n) = \sum_n^\infty m(A_n + t) \leq m^*(A+t) + \epsilon$

And

$m^*(A+t) \leq \sum_n^\infty m(A_n + t) = \sum_n^\infty m(A_n) \leq m^*(A) + \epsilon$

which then implies that $m^*(A+t) = m^*(A)$ as $\epsilon$ was arbitrary.

Dilation follows in a similar way: $A\subset \bigcup A_n \iff \delta A \subset \delta \bigcup A_n \iff \delta A \subset \bigcup \delta A_n$ , where $\delta > 0$ . Now, writing each $A_n$ as a finite union of disjoint intervals reveals