Given $A\subset [0,1]$ prove that $m^*(\{1-x:x\in A \})=m^* (A)$

Question

Given $A\subset [0,1]$ prove that $m^*(\{1-x:x\in A \})=m^* (A)$.

($m^*$ is outer Lebesgue measure)

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$$m^*(\{1-x:x\in A \})=\inf\{m^*(G):\{1-x:x\in A \}\subset G \mbox{ and } G \mbox{ is open}\}=\inf\{m^*(1-G):A\subset 1-G \mbox{ and } 1-G \mbox{ is open}\} \ge m^*(A)$$

I don't know how to prove the other inequality.

Answer 1

For $A \subseteq [0,1]$ let $S(A) = \{1-x \mid x \in A\}$. The statement is that $m^*(S(A)) = m^*(A)$, and you have already shown that $m^*(S(A)) \geq m^*(A)$.

Notice that $S^2(A) = A$ for every subset $A \subseteq [0,1]$, and thus $$ m^*(A) = m^*(S(S(A))) \geq m^*(S(A)) \geq m^*(A), $$ so we already have $m^*(S(A)) = m^*(A)$.