$m(\alpha E)=\alpha m(E)$ ? for every Lebesgue measurable set and $\alpha >0$

There are many ways of showing this.

One simple way is to take the linear transformation $f(x) = \alpha x$, and note that $\alpha E = f(E)$. Then the change of variables theorem gives $m (f(E)) = \int_{f(E)} 1 = \int_E (1 \circ f) J_f$, where $J_f(x) = |\det \frac{\partial f(x)}{\partial x}| = \alpha$. Since $1 \circ f = 1$, we have $m(F(E)) = \alpha \, m(E)$, as required.

Alternatively, you could note that if $\{I_k\}$ is a countable cover of $E$ by intervals, then $\{\alpha I_k\}$ is a countable cover of $\alpha E$ by intervals, ans similarly, if $\{J_k\}$ is a countable cover of $\alpha E$ by intervals, then $\{ \frac{1}{\alpha} J_k \}$ is a countable cover of $E$ by intervals. Since $l(\alpha I) = \alpha l(I)$ for intervals, it follows that the outer measure $m^*$ satisfies $m^* (\alpha E) = \alpha \, m^*( E)$, and hence it is true for $m$.

A third approach would be to note that the statement is true for intervals $I$, that is, $m(\alpha I) = l(\alpha I) = \alpha l(I) = \alpha m(I)$.

Any open set can be written as the disjoint union of a (well, at most) countable collection of intervals, hence if $U$ is open, we have $U = \cup_k I_k$, and $m (\alpha U) = \sum_k m(\alpha I_k) = \sum_k l(\alpha I_k) = \alpha \sum_k l(I_k) = \alpha \sum_k m(I_k) = \alpha m(U)$.

Finally, since $m$ is outer regular, for any measurable $E$ we have $m E = \inf \{ m(U) | E \subset U, \ \ U \text{ open} \}$. It follows that $m(\alpha E) = \alpha m(E)$ for any measurable set $E$.

Let $|I_n|$ be a covering sequence of open intervals such that $m^*(E) \le \sum |I_n| \le m^*(E) + \epsilon$ where $m^*$ denotes outer measure.

Write each $I_n$ as $(a_n,b_n)$. Define $\alpha I_n := ( \alpha \cdot a_n , \alpha \cdot b_n )$. Clearly, $\{ \alpha I_n \}$ is a covering sequence of intervals for $\alpha E$, since $\alpha \cdot e \in \alpha I_n$ for some $I_n$.

Moreover:

$$m^*(\alpha \cdot E) \le \sum | \alpha I_n | = \alpha \sum |I_n| \le \alpha m^*(E) + \alpha \epsilon$$

$\epsilon$ is arbitrary so $m^*(\alpha \cdot E) \le \alpha m^*(E)$.

The reverse inequality follows by the fact that $\alpha^{-1} (\alpha E) = E$, so $m^*(E) \le \alpha^{-1} m^*(\alpha \cdot E )$ by the exact same proof as above.

Let $M$ be the $\sigma$-algebra of measurable sets of the line and let $m:M\to\mathbb R$ be the Lebesgue measure.

The functions $\mu:E\in M\mapsto m(aE)\in\mathbb R$ and $\mu':E\in M\mapsto am(E)\in\mathbb R$ are $\sigma$-additive measures.

It is trivial to check that $\mu$ and $\mu'$ coincide on the subring of $M$ of intervals. Since $M$ is the completion of the $\sigma$-algebra generated by that subring, the usual theorems of extension of measures imply that $\mu=\mu'$.