TITLE: $m(\alpha E)=\alpha m(E)$ ? for every Lebesgue measurable set and $\alpha >0$
QUESTION [5 upvotes]: Let's consider a Lebesgue measurable set $ E \subset \mathbb R$. And let's consider a positive constant $\alpha>0$. I want to know if it's always true $m(\alpha E)=\alpha  m(E)$. Clearly $m$ denote the Lebesgue measure.
At least in the case where $mE=0$ or $\infty$ it's true and also in the case of $ E=[a,b]$ but I can't prove it or disprove it. Please help me. In the case that the result it's true I want to see some proof, if it's not, please I want to see a general family of sets that have that property.

REPLY [3 votes]: Let $M$ be the $\sigma$-algebra of measurable sets of the line and let $m:M\to\mathbb R$ be the Lebesgue measure.
The functions $\mu:E\in M\mapsto m(aE)\in\mathbb R$ and $\mu':E\in M\mapsto am(E)\in\mathbb R$ are $\sigma$-additive measures. 
It is trivial to check that $\mu$ and $\mu'$ coincide on the subring of $M$ of intervals. Since $M$ is the completion of the $\sigma$-algebra generated by that subring, the usual theorems of extension of measures imply that $\mu=\mu'$.