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I'm interested in potentially using the following definition to develop the basic theory of Lebesgue measurable sets, but I'm running into a considerable roadblock. First, some context

Definition: A set $A \subseteq \mathbb{R}^d$ is Lebesgue measurable if, for each $\varepsilon>0$, there exists an open set $O$ such that both $A \subseteq O$ and $m^*( O \setminus A) < \varepsilon.$

Note: $m^*$ is the Lebesgue outer measure. I find this definition appealing because closure under countable unions falls immediately out of it. So far, I've proven the following properties.

  1. If $A$ is open, then $A$ is Lebesgue measurable.
  2. If $A$ has outer measure zero, then $A$ is Lebesgue measurable.
  3. If $A = B \setminus C$ where $B$ is Lebesgue measurable and $C$ has outer measure zero, then $A$ is Lebesgue measurable.
  4. If $(A_n)$ is a sequence of of Lebesgue measurable sets, then $\bigcup_n A_n$ is Lebesgue measurable.
  5. If $A_1, \ldots, A_n$ are Lebesgue measurable, then $\bigcap_{i=1}^n A_i$ is Lebesgue measurable.
  6. If $A$ satisfies Caratheodory's criterion and $m^*(A) < +\infty$, then $A$ is Lebesgue measurable.
  7. If, for each $\varepsilon > 0$, there exists a closed set $F$ contained in $A$ such that $m^*(A \setminus F) < \varepsilon$, then $A^C$ is Lebesgue measurable.

I have not yet been able to prove any of the following, any one of which could be used to derive the other four.

  • If $A$ is Lebesgue measurable, then $A^C$ is Lebesgue measurable.
  • If $(A_n)$ is a sequence of of Lebesgue measurable sets, then $\bigcap_n A_n$ is Lebesgue measurable.
  • If $A$ is closed, then $A$ is Lebesgue measurable.
  • If $A$ is compact, then $A$ is Lebesgue measurable.
  • If, for each $\varepsilon > 0$, there exists a closed set $F$ contained in $A$ such that $m^*(A \setminus F) < \varepsilon$, then $A$ is Lebesgue measurable.

So, the question is can anyone enlighten me on how to prove any one of these five statements using only the above facts and some basic properties of the outer measure??

I've been stuck on this for a couple of days now. I've read every related question on this website which I could find (there are many, but they all take for granted that $A$ is LM iff $A^C$ is LM, which I have not yet been able to prove), and I've looked into several textbooks, but I can't make any headway.

I could really use either a massive hint, or affirmation that I'm engaged in a wild goose chase.

Thanks.

Edits: minor cosmetic changes

Thomas Winckelman
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    Well, I don't know if this will satisfy you, but it is a standard fact that your property is equivalent to measurability, so you could always prove it by passing through that fact. Also, if I am not mistaken, the last four properties on your list all follow from the first. –  May 25 '20 at 21:42
  • Yep, it certainly seems that if "$A$ is LM implies $A^C$ is LM," then everything else pretty much falls into place. I had intended to make that clear in the post. – Thomas Winckelman May 25 '20 at 21:47
  • I think I get what you're saying. I've seen a lot of good proofs that the two definitions are equivalent, but these proofs have all required that a lot of theory about Caratheodory's definition is first developed, and I guess what I'm truly wondering is if it's not easier to develop the theory from this direction. – Thomas Winckelman May 25 '20 at 21:52
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    Since this definition makes verifying some properties a lot easier, I would sort of "morally" feel like you'll have to pay the price for that simplicity somewhere else. Although the other tradeoff to this definition is that it doesn't generalize to arbitrary measure spaces, so maybe it really is a nice alternative for the real measure! –  May 25 '20 at 22:00
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    Take a look at Stein and Shakarchi's book on real analysis and measure theory. In chapter 1, they develop everything using this definition. (Although, I found Caratheodory's construction more systematic, and pretty intuitive as well) – peek-a-boo May 27 '20 at 02:02
  • Thanks for the recommendation! I haven't looked at that yet, and will do so now. – Thomas Winckelman May 27 '20 at 02:08

1 Answers1

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Short Answer

It appears that the standard approach is to first prove that closed sets are measurable in this sense, from which everything else on the list indeed follows. There are really only two kinds of proofs which I've seen, so far, of this fact.

  1. There is the proof of Property 4, Section 3, Chapter 1 of the book recommended by "peek-a-boo" in the comments. Actually, the same approach is taken by Terence Tao in this blog post.
  2. One can also simply prove directly that closed sets satisfy Caratheodory's criterion, and then use this to conclude that closed sets are measurable in the sense described above, as in Theorem 1.2, Section 1, Chapter 6 of the book recommended by "peek-a-boo."

To my mild chagrin, I find the second approach easier as well as more efficient, since in the process it also paves the way for a proof that the two definitions of "measurable" are equivalent.

Long Answer

This will still be open. If anyone viewing this at a later date knows of an elegant solution, I for one would love to hear about it. I can't help but imagine that there must be a more simple solution than those described above. The topology of open and closed sets has always been such a power tool, that I would be extremely surprised if it can't be used to solve the original question in a relatively direct manner...

Thomas Winckelman
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