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Title says it all, so I'll just repeat it: Does every Lebesgue measurable set have the Baire property?

2 Answers 2

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No. (assuming choice, so that there are some sets without the Baire property).

Some hints: Any set with the property of Baire which is not meager, has a subset that fails the property of Baire. Next, find a Lebesgue null set, with the property of Baire, but not meager.

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    @Theo: Ah! Thanks. That's what I've been missing all the time. The intersection isn't Q itself at all, but some larger set. And it's pretty easy to see that it's non-meager, too. Thanks folks!2011-09-07
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Just to make the solution from the comments to GEdgar's answer more explicit: The set of real numbers can be partitioned as $\mathbb R = M \cup N$, where $M$ is meager and $N$ is Lebesgue null. ($N$ is the dense $G_\delta$ described by t.b. above.)

Now every subset of $N$ is Lebesgue measurable, but (using AC) one can find a subset without the Baire property.

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    It's consistent that all sets have the Baire property but not all sets are measurable, though. So one has to use the axiom of choice here.2013-05-13