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It was a big curious for me that whether, for a given set $E\subset\mathbb R^p$ and an outer measure $\mu^*$, the existence of $\mu^*(E)$ implies the measurability of $E$. Plus, the same question if only limited to a Lebesgue measure $m$. Here, $\mu^*$ is denoted as $\mu$ for those sets whose measurability is already garanteed.

#1. According to the PMA(Principles of Mathematical Analysis) by Rudin, Chapter 11, the measurable set is defined as a countable union of 'finite measurable set' which is a set that has a sequence of elementary sets that converges to it (Here, elementary set just stands for the 'finite union' of intervals).

#2. According to Caratheodory, the set $E$ is called (Lebesgue) measurable if $$m^*(A) = m^*(A\cap E) + m^*(A\cap E^c)$$ for $\forall A\subseteq\mathbb R^p$.

Both definition does not intuitively gives clue that existence of outer measure of a given set implies its measurability. In addition, the use of expression $m^(A)$ in #2 makes feel more that there is a non-measurable set but still have an outer measure. There was a something interesting fact:

For a set $A$ that the outer measure exists(which might not be measurable in my sense now), there always exists a Borel set(which is measurable) $B$ and $C$ with $B\subset A\subset C$ such that $$\mu(B) = \mu^*(A) =\mu(C)$$

So, if we make an equivalent class $[A] = \{X: \mu^*(A\Delta X) = 0\}$ as suggested in PMA, then if $[A]$ has an outer measure, we can consider $[A]$ as a measurable set, in my opinion. Are these all logic can be justified?

Actually, these all criousity came up from the following question(this question was suggested based on the definition #1):

If $E$ has an outer measure $m^*(E)$, then for any open interval $(a, b)$, show that the following holds: $$b-a = m^*((a, b)\cap E) + m^*((a, b) \setminus E)$$

and having stuck in a moment that should $E$ be assumed to be measurable or not. I have also seen that the set $A$ in definition #2 may be limited to the intervals, i.e.,

#3. $E$ still can be called measurable if $$b-a = m^*((a, b)\cap E) + m^*((a, b) \cap E^c)$$ holds for all intervals $(a, b)$.

Consequently, if the bove equestion is actually true, then $E$ becomes a measurable set by #3. The question is very complicated and not very ordered, but so is my brain... Any help will be so appreciated.

ToBY
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  • It is not true that the existence of $\mu^* E$ implies measurability of $E$. – copper.hat Nov 07 '22 at 03:46
  • @copper.hat and are there any intuitive examples of such sets? I'd also like to know why the phrase "measurable" is used for such sets; even there is such couter-example. – ToBY Nov 07 '22 at 03:53
  • There are examples, of course, but they rely on the AOC so it is unlikely that intuition is involved. I doubt very much that the term measurable was used for non measurable sets, you should give a reference to such a statement. – copper.hat Nov 07 '22 at 03:58
  • If you want a response you need to use full sentences. – copper.hat Nov 07 '22 at 04:01
  • Sorry the comment has uploaded mistakenly, but could you refer some documents related to AOC, becuase I am not familiar with it. – ToBY Nov 07 '22 at 04:10
  • You need to so some work. https://math.stackexchange.com/questions/226559/examples-of-non-measurable-sets-in-mathbbr – copper.hat Nov 07 '22 at 04:13
  • For such non-measurable set, then, is it correct that definition #3 does not have to be always satisfied? – ToBY Nov 07 '22 at 04:18
  • Again, you need to do some work. It is a theorem that if the entire space $\Omega$ has finite outer measure then $E$ is measurable iff $\mu^* \Omega = \mu^* E + \mu^* E^c$. – copper.hat Nov 07 '22 at 04:25

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