1

Let $A \subseteq \mathbb{R}$ be Lebesgue measurable. Prove that: $\forall ε > 0, \exists F_ε ⊆ A: F_ε$ is closed in $\mathbb {R},$ & $λ^{∗} (A \smallsetminus F_ε) < ε$.

I have proven two similar results for an open subset version of this statement, got a logical equivalence of all of them in that case, and proved an analog of all of these for the closed version except this one. But I get hung up on this one. Primarily, it is due to the fact that I lose properties of either unions (countable unions of closed sets are not necessarily closed) or of the covering used in the definition of Lebesgue outer measure. I thought that I could try to develop a notion of inner measure (which I have not learned) and use finite unions, but I am not sure that I actually succeeded with the proof in the case of $λ(A) < + \infty$ (id est: the finite case); I was making up the notions via analogy and necessity and I am not sure that any of them were actually good. In any event, the finite union thing breaks down for the infinite case, so I am not sure how to fix my inner measure strategy.

I would prefer to just start over using outer measures, unions, and covers only. But I can make no progress. Do you have any good tips, at least to start it, for both the finite and infinite cases?

user332420
  • 171
  • I can provide examples of all of the work that I have done thus far, but I would rather not type it all up. I am slow at TeX. – user332420 Apr 18 '16 at 23:48
  • If you take an open cover of the complement, you get a closed subset of $A$. – Matt Samuel Apr 19 '16 at 00:45
  • Could you expound? I understand, but I tried something similar and could not get it. – user332420 Apr 19 '16 at 01:15
  • First restrict your attention to a closed interval. The suggestion should lead to a proof in that case. Then extend by taking the union. I think you can prove in this case that the set remains closed if you're taking the union over intervals that intersect in at most their endpoints. – Matt Samuel Apr 19 '16 at 02:13

0 Answers0