$\mu^{\ast}(\alpha A) = |\alpha|^2 \mu^{\ast}(A)$ I'm trying to prove that this statement in real analysis:

If $A \subset R^2$ is measurable, then $\alpha A$ is measurable and

$$\mu^{\ast}(\alpha A) = |\alpha|^2 \mu^{\ast}(A)$$ 

Thanks for your help and time.

My idea is this:

$$
m^*(E)=\inf\left\{\sum_{n≥1}\ell (I_n)\middle|E\subset\cup_{n≥1}I_n\right\}
$$
and we note that $m(E)=m^*(E)$ for measurable sets $E$.

Noe note that any set $A\subseteq(a,b)$ satisfies $\alpha A\subseteq(\alpha a,\alpha b)$ and that $\ell(\alpha(a,b))=\alpha \, \ell(a,b)$.  It follows that for any collection of sets $\{I_n\}_{n≥1}$ whose union contains $E$, we may state that $\{\alpha I_n\}_{n≥1}$ is a collection whose union contains $\alpha E$ and whose total length is $\alpha$ times longer.  For any set $A$ of real numbers, $\inf (\alpha A) = \alpha \inf A$.  The desired statement follows.

Is this correct?
Use the uniqueness of the Lebesgue measure.  The Lebesgue measure $\mu$ on $\mathbf R^n$ is the unique measure on the Borel sets of $\mathbf R^n$ satisfying the following properties:

1 .  $\mu(E + x) = \mu(E)$ for all Borel sets $E$ and all $x \in \mathbf R^n$.
2 .  Every Borel set can be measured from within by compact sets, and measured from without by open sets.
3 .  $\mu( [0,1] \times \cdots \times [0,1]) = 1$.
Use the uniqueness of the Lebesgue measure.  The Lebesgue measure $\mu$ on $\mathbf R^n$ is the unique measure on the Borel sets of $\mathbf R^n$ satisfying the following properties:

1 .  $\mu(E + x) = \mu(E)$ for all Borel sets $E$ and all $x \in \mathbf R^n$.
2 .  Every Borel set can be measured from within by compact sets, and measured from without by open sets.
3 .  $\mu( [0,1] \times \cdots \times [0,1]) = 1$.