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Is there any Characterization for all measurable sets in $\mathbb{R}$? Can I say that a set is measurable if an only if it has the property of Baire? (differs from an open set by a first category set). If the answer is no, Is there an example of a measurable set that does not have the property of Baire?

Thank you! Shir

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What about Lebesgue regularity? Any measurable set is a countable intersection of open set, then subtract away a null set. Or are you looking for a different kind of characterization? –  Gina Dec 29 '13 at 17:51
    
Anything that will approve my understanding. Your characterization helped. Thank you! –  Shir Sivroni Dec 29 '13 at 21:32

2 Answers 2

up vote 5 down vote accepted

A useful characterization of measurable sets is that a set is measurable if and only if it is of the form $B\triangle N$ where $B$ is Borel (even $G_\delta$) and $N$ is null (measure zero) (this is easy to show using regularity of the measure, and is actually true for any regular Borel measure, on any space).

From that it's not hard to see that a measurable set can be rather pathological: $N$ can be any subset of the Cantor set, for instance. It can also be any subset of a comeager set which is of zero measure (like the set of Liouville numbers, for example), which can fail to have Baire property. It also makes it easy to see that there are $2^{\mathfrak c}$ measurable sets.

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Thank you tomasz! –  Shir Sivroni Dec 29 '13 at 21:30

To answer your second question, it is not true that a set is measurable iff it has the Baire property.

There is a way to see how measurable subsets $U\subset \mathbb{R}$ look like though, it is called Littlewood's first principle and states:

If the outer measure $\mu^*(U)<\infty$ then $U$ is measurable if and only if for all $\epsilon > 0$ there is a finite union of intervals $J$ such that $\mu^*(U \triangle J)< \epsilon$.

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I see. will have to think about it. Thank you! –  Shir Sivroni Dec 29 '13 at 21:31

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