I will show an version of Arzela-Ascoli theorem, which is called Kolmogorov-Riesz theorem, by using a fact in topology as following.

DefinitionIn a metric space, a set is totally bounded, if for any fixed , the set can be covered by finitely many open balls of radius .

LemmaIn a metric space, a set is compact if and only if is complete and totally bounded.

Proof of the lemma can be find here.

Theorem 1 (Kolmogorov Riesz Theorem (finite measure domain))Let be a bounded set of functions, Assume that , there exists a such thatwhenever , for all . Then has a compact closure in for any finite measure set .

*Proof:* Since is complete, by the above lemma we need to show is totally bounded. The idea is first find a bounded domain with almost same measure, use Arzela-Ascoli theorem to get compactness for , then we have finite open balls covering . Enlarge those balls a bit to get a finite cover for .

Let be a sequence of mollifiers with supp. Fix , also fix such that , we have

, for any . Direct verification shows for any ,

which implies

By (2) and the inner regularity of measurable sets, we can find a bounded measurable set such that . Therefore

By (2) and (3) we know the functions in the set are bounded and equicontinuous, so we can apply Arzela-Ascoli theorem and hence has a compact closure in . This means for every sequence in , there exists a uniformly convergent subsequence. Since the subsequence also converges in , we know that has a compact closure in . Therefore, can be covered by finite ball of radius in , say . i.e.

, where . Now define

Finally I claim that . For any ,

Choose such that , along with the estimate (1) and (4), we have

This proves that is totally bounded, hence has a compact closure in .

Theorem 2 (Kolmogorov Riesz Theorem (infinite measure domain))Let be a bounded set of functions, Assume that , there exists a such thatwhenever , for all . Furthermore, , there exists an with finite measure such that

Then has a compact closure in .

*Proof:* By the previous theorem, fix , . Define

Then by condition (5) and similar calculation in the previous theorem, .

RemarkThe condition (5) for infinite measure domain is essential. A counterexample without (5) would be as following.Let be any function, with supp. Let , and . So is a bounded set,whenever . However,

if . This implies does not have any convergent subsequence. Hence does not have compact closure in .