# 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? Posted on