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If the answer is yes, how to prove that? Otherwise how to find a counterexample?

Update:

I've figured out the tricks inside.

A countable intersection of open sets in $\mathbb R$ is equivalent to a countable union of closed sets. Since a countable union of a sequence of pairwise disjoint measurable sets is again measurable, will have a countable union of a sequence of any measurable sets measurable. Since a closed set is measurable, apparently, a countable intersection of open sets in $\mathbb R$ is measurable.

P.S. I'm not sure why this question is off-topic. However, it will help a real variable novice make clear sense of the Borel σ-algebra and the Borel sets.

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    $\begingroup$ its complement is a countable union of closed sets $\endgroup$ – Rolf Hoyer Apr 17 '15 at 22:57
  • $\begingroup$ @RolfHoyer: Yes. $\endgroup$ – Bear and bunny Apr 17 '15 at 22:59
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Sigma algebras are closed under countable intersections. Combine this with the fact that Borel measurable sets are also Lebesgue measurable is sufficient to answer your question.

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