I want to be sure that any subset of a measurable set of finite measure has a finite covering.
This question is because, I have seen that for any measurable set E of finite measure, for each $\epsilon > 0$ we can find an open set $O$ which is a finite union of disjoint intervals and covers the set $E$, with $m(O\setminus E) < \epsilon$. That means the difference between the set and the open set is less than epsilon for any epsilon and the set $O\setminus E$ have measure 0. But in some proofs, I have seen that even when we the set in which we are working is compact (ex: [0,1] or [a,b]) they only assume the existence of the countable covering instead of the finite. Why? is that necessary? By Heine-Borel I know that every open covering has a finite subcover, so why complicate the problem with countable coverings when is easier to prove it by finite covers?.