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So from Caratheodory's condition we can show that a set is measurable if it can be enclosed in an open set whose measure is equal / arbitrarily greater than the original set.

What is the reason that we are only considering outer approximations by open sets, and inner approximations by closed sets?

Thanks.

user49404
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  • Is this reasoning correct? Because the fat Cantor set (Smith-Volterra-Cantor set) is a counterexample in both conditions. The complement of the fat Cantor set can't be outer-approximated by closed sets arbitrarily closely because its closure has non-zero measure, and the fat Cantor set can't be inner-approximated by open sets arbitrarily closely because it has non-zero measure and contains no interval. – user49404 Dec 14 '18 at 22:40

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Note that points are closed, so $\bigcup_{x\in X}\{x\}$ is always a covering of $X$ by closed sets (regardless of what $X$ is); since we want points to have measure zero, this suggests that every set has measure zero, which is clearly undesirable.

However, it turns out that if we use closed intervals (or balls in higher dimensions) of positive radius instead of open intervals/balls, we do indeed get the same theory, and this is a good exercise.

Noah Schweber
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