Does an isometric transformation preserve the Lebesgue outer measure? Let $T:\mathbb{R}\rightarrow\mathbb{R}$ be an isometric function i.e.
$$|x-y|=|T(x)-T(y)|$$
for all $x,y\in \mathbb{R}$
Consider now Lebesgue outer measure:
$$\mu^*(A)  := \inf\{\sum _{i\in I}|P_i|:A\subseteq \bigcup_{i\in I} P_i\}$$
Wherer $I$ is countable set and $P_i$ is an open interval.

Does equality $\mu^{*}(A) = \mu^{*}(T(A))$ holds?

I am only able to see that for open intervals we have $|P|=|T(P)|$. I have an intuition that this is somehow sufficient to solve the problem.
Thank you in advance for help.
 A: There is a theorem that any isometry of $\mathbb R^n$ is affine. Hence, you actually have $T(x)=x+c$ or $T(x)=-x+c$, and your proof becomes much simpler:
Now you can use that actually $T(P)$ is an open interval whenever $P$ is, and conversely if $Q$ is an open interval then so is $P=T^{-1}(Q)$ with $T(P)=Q$. Depending on your definition of outer measure the claim is now trivial.
A: You are right! In fact, as a general principle (which can be made precise) you can think that any operation that does not change the measure of open intervals, would not change the measure or outer measure of any set.
For your particular problem, you already know that $|P|=|T(P)|$ for every open interval $P.$ Another little fact that you might want to prove is that if $P$ is an open interval, then $T(P)$ is also an open interval.
Now fix $A\subseteq \mathbb{R}$ such that $\mu^*(A)<\infty.$ This means that for any $\epsilon>0,$ you have a collection of open intervals $(P_{i})_{i\in I}$ such that $A\subseteq \bigcup_{i\in I} P_i$ and $$\sum_{i} |P_i|\ge \mu^*(A)\ge \sum_{i} |P_i|-\epsilon.$$
Argue that $T(A)\subseteq \bigcup_{i\in I}T(P_i).$ Using the fact that $T(P_i)$ is open intevral, it follows that
$$\mu^*(T(A))\le \sum_{i\in I}|T(P_i)|=\sum_{i\in I} |P_i|\le \mu^*(A)+\epsilon.$$
Since $\epsilon$ was arbitrary, it follows that $\mu^*(T(A))\le \mu^*(A).$
Now use the fact that if $T$ is an isometry then $T^{-1}$ can be defined and it is also an isometry. Using this and repeating the above argument exactly one obtain $\mu^*(A)\le \mu^*(T(A)).$
