Proving $m(E)<\infty\iff \forall\epsilon>0\exists\text{ compact }F\subset E: m(E)-m(F)<\epsilon$

Question

Is proving this question:

Proving $m(E)<\infty\iff \forall\epsilon>0\exists\text{ compact }F\subset E: m(E)-m(F)<\epsilon$.

The same as proving this question:

Let $E\subset \mathbb{R}$ be a measurable set of positive measure, and let $0 < \alpha < m(E)$ Prove that there exists a compact set $F\subset \mathbb{E}$ such that $m(F) = \alpha$.

If yes, why? if no, also why? What will be the difference in proving the second one as I know the proof of the first one?

Thanks!

Answer 1

Let $K$ be a compact subset of $E$ such that $\alpha < m(K)<m(E)$. Prove that $f(x)=m((-\infty, x]\cap K)$ is a continuous function and use IVP to prove the second result.

$f(y)-f(x)=m((x,y]\cap K) \leq m((x,y])=y-x$ for $x<y$ and this proves that $f$ is continuous. Also $f(x) \to 0$ as $x \to -\infty$ and $f(x)\to m(K)$ as $ x\to \infty$. Now IVP tells you that $f$ attains all values between $0$ and $m(K)$, in particular the value $\alpha$.

Note that $(-\infty, x]\cap K$ is a closed subset of $K$ and hence it is compact.