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 $E,F\subset \mathbb{R}$



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 $E=\mathbb{Q}$.


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.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.