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

Suppose $E$ is measurable. Show that $m(E)<\infty\iff \forall\epsilon>0\exists\text{ compact }F\subset E: m(E)-m(F)<\epsilon$

where $m$ is Lebesgue measure, and $E,F\subset \mathbb{R}$

Attempt/Thoughts:

$(\implies)$

Suppose $E$ is measurable and $m(E)<\infty$. Then there exists a closed set $F\subset E$, such that $m(E\setminus F)<m(E)-m(F)<\epsilon$. I'm not sure how to show that $F$ is bounded though (since if it was bounded, then it would be both closed and bounded and hence compact by Heine-Borel). At first I thought $m(E)<\infty$ would mean that $E$ is bounded, but then I remembered that even if $m(E)<\infty$, that doesn't necessarily mean that $E$ is bounded, which we can see if you consider $E=\mathbb{Q}$.

$(\impliedby)$

Suppose $E$ is measurable, and $\forall\epsilon>0\exists\text{ compact } F\subset E: m(E)-m(F)<\epsilon$.

Not sure how to continue in this direction at all.

Any help would be appreciated. Thanks.

• Is everything in $\mathbb R$ and $m$ is the Lebesque measure? – user99914 Feb 9 '15 at 5:25
• Yes, that's correct. – Sujaan Kunalan Feb 9 '15 at 5:26

Hint: ($\Rightarrow$) Define $E_n = E \cap [-n, n]$. Then $m(E_n) \leq m(E_m)$ if $m\geq n$ and

$$\lim_{n\to \infty} m(E_n) = m(E) < \infty$$

So there is $n_0$ so that

$$m(E_{n_0}) \geq m(E) - \epsilon/2 .$$

Now this $E_{n_0}$ is bounded.

($\Leftarrow$) This is easier: $m(E) < m(F) + \epsilon$. But $F$ is compact....

• $F$ is compact, so its measure must be finite, and so $m(E)$ is bounded. Of course! Thanks. – Sujaan Kunalan Feb 9 '15 at 5:45