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I am reading the final chapter of baby rudin. I have trouble understanding the difference between $\mu$-measurable sets and the Borel sets. The remarks on page 309 imply that the latter is contained in the former. But from the way I see it, they are both the collection of sets, which are the final forms of open intervals after undergoing intersections, taking complements and unions. So what exactly are $\mu$-measurable sets? And why are the Borel sets $\mu$-measurable for every $\mu$?

XXX
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  • Welcome to mse! I'm not sure I understand your question. Are you asking why every borel set is $\mu$-measurable (where I assume $\mu$ is lebesgue measure?)? Or are you asking why there are $\mu$-measurable sets that are not borel? – HallaSurvivor Mar 06 '21 at 04:13
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    So yes, all Borel sets are measurable. There should be an example somewhere of a set defined using the Cantor function of a measurable set that is not Borel. – ndhanson3 Mar 06 '21 at 05:22
  • Also, $\mu$ is not a variable, $\mu$ is shorthand for the Lebesgue measure. – ndhanson3 Mar 06 '21 at 05:23
  • @HallaSurvivor Hi ! I think $\mu$ can be any measure, not just the lebesgue measure. I am very confused about the concept of $\mu-measurable$ sets. Are they defined for any measure, i.e.a collection of measurable sets or they only refer to one particular measure? – XXX Mar 06 '21 at 06:10
  • @ndhanson3 I think Rudin uses $m$ for the Lebesgue measure and $\mu$ for any kind of measure – XXX Mar 06 '21 at 06:10
  • This question has the construction but it's involved. It is hard to find borel sets which aren't lebesgue measurable but they exist. – Daniel Mar 06 '21 at 07:32
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    @DanielApsley You mean Lebesgue measurable but not Borel. – ndhanson3 Mar 06 '21 at 07:53

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For any measure $\mu$, remark (a) says that the open sets are $\mu$-measurable. A set $E$ is $\mu$-measurable simply if $\mu(E)$ is defined. Not every set is measurable though.

So the set of measurable sets $\mathfrak{M}(\mu)$ is a $\sigma$-algebra containing the open sets. Since the Borel set is the smallest $\sigma$-algebra containing the open sets, $\mathfrak{M}(\mu)$ must contain all the Borel sets. This is indeed regardless of what measure we start with.

ndhanson3
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