How to show that the set $aE=\{ax: x \in E\}$ is Lebesgue measurable? 
Let E be a Lebesgue measurable set. Show that $aE=\{a*x: x \in E\}$ is
  Lebesgue measurable.

My attempt:
If $a =0$, then $aE$ is either only the singleton consisting of $0$, or the empty set. Both are known to be measurable. If a is not zero I have to show that:
$\lambda^*(W)=\lambda^*(W\cap aE)+\lambda^*(W\cap (aE)^c)$, this is the caratheodiry criterion. And the function is the lebesgue other measure. I have to show that it holds for any W, using that E is measurable. The trick is the write $W\cap aE=S_1\cap E$, where $\lambda^*(W\cap aE)=\lambda^*(S_1\cap E)$. 
And $W\cap (aE)^c=S_2\cap E^c$, where $\lambda^*(W\cap (aE)^c)=\lambda^*(S_2\cap E^c)$. And then use the measurability of E.
However, I don't know how to find $S_1$, and $S_2$?
 A: Let $\frak m^*$ be the Lebesgue outer measure. Remember that ${\frak m}^*(aE) = |a| {\frak m}^*E$. If $a = 0$, it is trivial. Suppose $a \neq 0$. Let $A \subset \Bbb R$ (or $\Bbb R^n$, whatever) be any set. Consider the set $\frac{1}{a}A$. Since $E$ is measurable, we have that: 
$$ {\frak m}^*\left(\frac{1}{a}A\right) = {\frak m}^* \left(\frac{1}{a}A \cap E\right) + {\frak m}^*\left(\frac{1}{a}A \cap E^c\right) $$
Multiply it all by $|a|$. We get: $$ |a|{\frak m}^*\left(\frac{1}{a}A\right) = |a|{\frak m}^* \left(\frac{1}{a}A \cap E\right) + |a|{\frak m}^*\left(\frac{1}{a}A \cap E^c\right) \\  {\frak m}^*A = {\frak m}^* \left(a\left(\frac{1}{a}A \cap E\right)\right) + {\frak m}^*\left(a\left(\frac{1}{a}A \cap E^c\right)\right) $$
Before we go on, notice that $\alpha(X \cap Y) = \alpha X \cap \alpha Y$ and $(\alpha X)^c = \alpha X^c$ for any sets $X,Y$, and $\alpha \in \Bbb R$. Bearing this in mind, we go on:
$$ {\frak m}^*A = {\frak m}^* \left(a\frac{1}{a}A \cap aE\right) + {\frak m}^*\left(a \frac{1}{a}A \cap aE^c\right)  \\ {\frak m}^*A = {\frak m}^* \left(A \cap aE\right) + {\frak m}^*\left(A \cap (aE)^c\right) $$
and hence $aE$ is measurable.
A: If you notice that there is a one-to-one correspondence between open covering by intervals between a set $F$ and $aF$, via $(x,y)\mapsto (ax,ay)$, you can deduce that $\lambda^*(aF)=|a|\lambda^*(F)$ for any set $F$. 
Then, after checking that $W\cap aE=a(a^{-1}W\cap E)$ and $W\cap (aE)^c=a(a^{-1}W\cap E^c)$,
$$
\lambda^*(W\cap aE)=\lambda^*(a(a^{-1}W\cap E))=|a|\,\lambda^*(a^{-1}W\cap E).
$$
Similarly, $\lambda^*(W\cap(aE)^c)=|a|\,\lambda^*(a^{-1}W\cap E^c)$. So
$$
\lambda^*(W\cap aE)+\lambda^*(W\cap(aE)^c)=|a|\,\left[\lambda^*(a^{-1}W\cap E)+\lambda^*(a^{-1}W\cap E^c)\right]=|a|\,\lambda^*(a^{-1}W)=\lambda^*(W).
$$
