# Lebesgue measure: show that $\mu^*(k\,A)=k \mu^*(A)$

For $k\gt 0$ and a subset $A$ of $\mathbb{R}$, let $k\,A=\{kx\mid x∈A\}$. Show that $$\mu^*(k\,A)=k \mu^*(A)$$ and that $A$ is measurable if and only if $k\,A$ is measurable.

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assuming A is measurebale, I tried to use the outer measure definition.And I know that, if a set A is covered by an interval [a, b], then the corresponding set kA is covered by the interval [ka, kb].but I can't recall any path to find the answer for this. – ccc Oct 21 '12 at 15:08
Denote by $l([a,b])$ the length of the interval $[a,b]$. Is it clear that $l([ka, kb])= k\cdot l([a,b])$? – Giovanni De Gaetano Oct 21 '12 at 15:13

To show the equality, you can try showing each is $\leq$ the other by using the definition of outer measure, i.e. let $A$ be covered by a countable union of intervals $I_{n}$ such that $\mu^{*}(A)+\varepsilon<\sum_{n}l(I_{n})$, then go on to say something about $\mu^{*}(kA)$, and vice versa.
Take a subset $B$ of $\mathbb{R}$. Then by the equality to be shown, you can get $\mu^{*}(B\cap A)+\mu^{*}(B\cap A^{c})=\mu^{*}(k(B\cap A))+\mu^{*}(k(B\cap A^{c}))=\mu^{*}(kB\cap kA)$ $+\mu^{*}(kB\cap kA^{c})$ which may be useful for the second statement.