Let $A \subset \mathbb{R}^n, k> 0$. Define $kA = \{kx : x \in A \}$. Show that $m^\ast(kA)=km^\ast(A).$ 
Let $A \subset \mathbb{R}^n, k> 0$. Define $kA = \{kx : x \in A \}$. Show that $m^\ast(kA)=km^\ast(A).$

Assuming that $\mathcal{U}$ is a Lebesgue cover for $A$, then I'm first trying to show that $\mathcal{U_k} = \{kI : I \in \mathcal{U} \}$ is a Lebesgue cover for $kA$. Pick $a \in kA$, then $a$ is of the form $$a=ka', \text{where $a' \in A.$}$$
Now since $\mathcal{U}$ is a cover for $A$ there exists $U \in \mathcal{U}$ such that $a' \in U$. Now $$a=ka' \in kU \in \mathcal{U_k}.$$ And thus $a \in \bigcup \mathcal{U_k}$ which would imply that $\mathcal{U_k}$ is a cover for $kA.$ Now it seems that we need to prove that $\mathcal{U_k}$ is generated by open intervals? If this is the case then pick $I = (a_1,b_1)\times(a_2,b_2)\times \dots \subset \mathbb{R}^n$. Now $$x \in kI \iff \frac1kx \in I \\  \iff \frac1kx \in (a_1,b_1)\times(a_2,b_2)\times \dots \\ \iff x \in (ka_1,kb_1) \times (ka_2,kb_2) \times \dots$$
So the cover $\mathcal{U_k}$ is indeed generated by open intervals. Now I'm left wondering how is this gonna help me show $m^\ast(kA)=km^\ast(A)$? If $$m^\ast(kA)= \inf\{\sum_{j=1}^\infty m(I_j) : kA \subset \bigcup_{j=1}^\infty I_j \}$$
then how can I show that this equals $km^\ast(A)$?
 A: As you showed, covers of $A$ give covers $U_k$ of kA. This shows that
$$m^*(kA) \leq km^*(A)$$
To show that $m^*(kA) \geq km^*(A)$ just do the opposite: any cover of $kA$ can be scaled by $\frac{1}{k}$ to give a cover of $A$.
Edit: The fact that $m^* (kA) \leq k m^* (A)$ is because if $\mathcal{U}$ covers $A$ then $k\mathcal{U}$ (you called this $\mathcal U_k$) covers $kA$. Therefore for any cover $\mathcal U$ of $A$,
$$m^*(kA) \leq \sum_{U \in \mathcal U} k m(U) = k \sum_{U \in \mathcal U}  m(U)$$
since $k \mathcal U$ is a cover of $kA$. Since this is true for all $U$, it is true of the infimum (by definition of infimum):
$$m^*(kA) \leq \inf_{\mathcal U} \sum_{U \in \mathcal U} k m(U) = k \inf_{\mathcal U} \sum_{U \in \mathcal U}  m(U) = k m^*(A)$$
A: Hint
It follows immediatelly from your work that for all $A,k$ you have
$$
m^*(kA) \leq km^*(A)
$$
In particular, replacing $k$ by $\frac{1}{k}$ and $A$ by $B=kA$ you get
$$
m^*(\frac{1}{k}B) \leq \frac{1}{k}m^*(kA)
$$
What is $\frac{1}{k}B$?
