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I am studying elementary measure theory from the book Real Analysis by Stein & Shakarchi.

The book defines $m_{*}(E)$ as the outer measure of some set $E \subset \mathbb{R}^d$ and presents the property:

If $E \subset \mathbb{R}^d$, then $m_*(E) = \inf_{E \subset \mathcal{O} \text{ open }} m_*(\mathcal O)$.

Moreover, $E \subset \mathbb{R}^d$ is measurable if for any $\epsilon > 0$ there exists an open $\mathcal{O}$ such that $E \subset \mathcal O$ and $m_{*}(\mathcal O - E) \leq \epsilon$.

I find it hard to intuitively understand the difference between the property above and the definition of measurable sets. If, for example, the property was satisfied with a $\min$ instead of $\inf$ would it then imply that any set in $\mathbb{R}^d$ is measurable? In general, how can I visualize a set such that its outer measure is the same as the infimal outer measure of the open sets containing this set, but there does not exist an open set containing this set such that the outer measure of the difference can be bounded arbitrarily?

independentvariable
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    @JoséCarlosSantos thanks for showing this to me. I think the definitions are slightly different (the other questions seem to be answering this intuition about Caratheodory-measures / extension theorem). Also, in my question, my main motivation is to intuitively understand the difference between the property I shared, and whether having min instead of inf would result in equivalent definitions. If you think this is irrelevant I can delete my questoin. – independentvariable May 03 '21 at 15:11
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    @independentvariable: The definition José is referring to and the one in your OP are equivalent, but some work is needed to show that. In any event, the inntiuition is that measurable sets should be objects that one can approximate by well known sets (open) whose measures one "knows" how to estimate. Besides that, the idea is ore about ingenuity, for it puts things on solid ground. – Mittens May 03 '21 at 15:15
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    Have you studied an example of a non-measurable set yet? That will probably help, as you'll be able to see why, for such a set $E$, $m_*(\mathcal{O} - E)$ cannot be made small. – Nate Eldredge May 03 '21 at 15:41
  • I don't understand what you mean with your question about replacing $\inf$ by $\min$. For many simple sets, the minimum is not attained; take for instance $E = [0,1]$ in $\mathbb{R}^1$. So if you try to define outer measure with $\min$, then the outer measure of such sets would simply not exist. – Nate Eldredge May 03 '21 at 15:44
  • @NateEldredge of course min would not be true. I was just wondering if the property always holds with min would this imply any set is measurable. – independentvariable May 03 '21 at 15:45
  • Sorry, I am still confused. You say "if the property always holds with min", when it obviously doesn't always hold with min; it sounds like asking "If pigs could fly...". Maybe I am misunderstanding what you mean; could you write out a precise statement, making the quantifiers clear? – Nate Eldredge May 03 '21 at 15:49
  • @NateEldredge Exactly! What I ask is something like 'If pigs could fly, does this imply pigs are birds'. In other words, if the property was true with a min, then any set would be measurable, as we could then take $\mathcal O$ as the minimizer in the first definition. Sorry if this is confusing, I will edit. – independentvariable May 03 '21 at 15:51
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    I see. Maybe a better way of saying is that "every set $E$ for which the min is attained is a measurable set". It's not true, though. The usual Vitali set $V$ is a non-measurable subset of $(0,1)$ having outer measure 1, so the inf in the definition of $m_*(V)$ is attained by taking $\mathcal{O}=(0,1)$. – Nate Eldredge May 03 '21 at 15:55
  • @NateEldredge thanks for your time, and sorry for my bad statements. Measure theory is still hard for me to imagine and proces.s – independentvariable May 03 '21 at 15:57

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