Prove: $m^*(aE+b)=|a|m^*(E)$

Question

For all $E\subseteq \mathbb{R} $ and $a,b\in \mathbb{R}$ we define $aE+B:=\{ax+b:x\in E\}$

Prove: $$m^*(aE+b)=|a|m^*(E)$$

I know the following properties:

1. $m^*(\emptyset)=0$ and so does for $m^*(r)=0$ when $r\in \mathbb{R}$

2. $A\subseteq B \Rightarrow m^*(A)\leq m^*(B)$

3. $m^*(A\cup B)\leq m^*(A)+m^*(B)$

Now we can say that if $aE$ is covered by $\{I_n\}: \bigcup_{i=1}^{\infty}I_n$ so $aE+b$ is covered by $\bigcup_{i=1}^{\infty}I_n+b$ so from the definition of an exterior mesure ($m^*(E)=|E|$) and property $1$ we get $$m^*(aE+b)=m^*(aE)$$

I can I prove that $m^*(aE)=|a|m^*(E)$?

Answer 1

Here's how I would do it: Let $\epsilon>0$ and assume $a\neq 0.$ Then,

$1).\ $ Let $(I_n)$ be a sequence covering $E$ such that $m^*(\sum |I_n|)\le m^*(E)+\epsilon.$

$2).\ $ If $I_n=(a_n,b_n),\ $ then $aI_n=(aa_n,ab_n)$, so $(aI_n)$ covers $aE$ and

$3).\ m^*(aE)\le m^*(\sum |aI_n|)=am^*(\sum |I_n|)\le am^*(E)+a\epsilon$, so

$4).\ m^*(aE)\le am^*(E).$

$5).\ $ On the other hand, the reverse inequality comes free, because if we replace $E$ by $aE$ and repeat the steps $1).-4).$, we have $\frac{1}{a}(aE)=E\Rightarrow m^*(E)\le \frac{1}{a}m^*(aE)\Rightarrow m^{*}(aE)\ge am^{*}(E)$.