Show that $\mu(aE)=|a|\mu(E)$

Question

Let $E\subset\mathbb{R}$ Lebesgue measurable. First of all, is $aE=\{ax:x\in E\}$ measurable for a real number $a$?

Second, how should I tackle the problem of showing that $\mu(aE)=|a|\mu(E)$ where $\mu$ is the Lebesgue measure?

Answer 1

Guide

Define the measure $\nu$ on the Lebesgue measurable sets by $\nu E=\mu(aE)$. Show that $\nu$ and $|a|\mu$ agree on the algebra of finite disjoint union of rectangles and hence by the Caratheodory extension theorem agree on the Borel sets. To show that they agree on the Lebesgue measure sets recall that any Lebesgue measurable set $F$ may be written as $F=B\cup N$ where $N$ is a null set and $B$ is a Borel set. Show that $aF=aB\cup aN$ and that $aN$ is a null set and bootstrap the previous results to show that $\nu F=|a|\mu F$.

Answer 2

I will assume $a\neq 0$, as the $a=0$ case is easy.

1. $aE = f^{-1}(E)$, where $f(x)=x/a$ is a measurable map.

2. First, prove that this is true in the special case that $E$ is a half-open interval, $(b,c]$. Then, use the fact that $$\mu(E)=\inf \sum_i \mu(I_i),$$ where the infimum ranges over countable lists of intervals whose union covers $E$, to extent this to all sets.
   
   In more detail: since $aE\subset \bigcup_i I_i$ if and only if $E\subset \bigcup_i a^{-1}I_i$, $$\mu(aE)=\inf_{aE\subset \bigcup_i J_i} \sum_i \mu(J_i)=\inf_{E\subset \bigcup_i a^{-1}J_i} \sum_i \mu(a\cdot a^{-1}J_i)=\inf_{E\subset \bigcup_i I_i} \sum_i |a|\mu(I_i)=|a|\mu(E)$$In the third inequality, we "reindex" the sum by letting $I_i=a^{-1}J_i$. In that step, we simultaneously use the fact $\mu(aI_i)=|a|\mu(I_i)$, which was proved before.