How will m*(rE) behave?

Question

Let $rE ＝\{rx: x\in E\}$, what is $m^*(rE)$ in terms of $m^*(E)$?

Intuitively, I think $m^*(rE)\leq r\times m^*(E)$. However I've no idea how to prove it?

Add: Definition of Lebesgue Outer Measure is here: How to explain the why here?

Update: My intuitiveness can be incorrect. Actually, $m^*(rE) = r\times m^*(E)$. Still, I've no idea how to find a complete proof.

Answer 1

You are right (as long as the underlying space is $\mathbb R$) but you get the reverse inequality more or less for free: $$ m(E) = m \left(\frac 1r rE \right) \le \frac 1r m(rE)$$ so that $rm(E) \le m(rE)$.