measure theory – Property of dilatation in sets measureble Lebesgue


I need to prove this problem:

Problem: Prove that $$forall alpha in mathbb{R}, forall E in mathfrak{M}_{mu}(mathbb{R}): mu(alpha E)=|alpha|mu(E)$$

Definition $1$: $$E subseteq mathbb{R} quad text{is measurable Lebesgue} quad iff forall epsilon>0, exists G quad text{abierto}: Esubseteq G quad wedge quad mu^{*}(Gsetminus E)leq epsilon $$

Definition $2$: $$mathfrak{M}_{mu}(mathbb{R})={E subseteq mathbb{R}: E quad text{measurable Lebesgue $mu$}}$$

Notation: Here $mu^{*}$ denote outer measurable and $mu$ denote measurable Lebesgue.

My attempt: First, we need to prove that $alpha E$ is measurable Lebesgue. For that, we need to considerer two cases:

  1. $alpha=0$.
    Proof:
    Let $alpha=0$ and $Ein mathfrak{M}_{mu}(mathbb{R})$, so we can see by definition that $$0E={0cdot x: xin E}={0}in mathfrak{M}_{mu}(mathbb{R}).$$
  2. $alphanot=0$.
    Proof:
    Let $alphanot=0$ and $E in mathfrak{M}_{mu}(mathbb{R})$, we can prove that that $$alpha E={alpha cdot E: xin E} in mathfrak{M}_{mu}(mathbb{R})$$but, $alpha E$ is measurable Lebesgue if, and only if, $forall epsilon>0$, we can prove that there exists a open set $G$ such that $$alpha Esubseteq G quad wedge quad mu^{*}(Gsetminus alpha E)leq epsilon$$
    First question: How can I prove this part?