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

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2 Answers 2

up vote 4 down vote accepted

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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Care to give an additional hint on the latter (i.e. Lebesgue null but not meager)? I can find non-meager sets with arbitrarily small Lebesgue measure, but not with measure = 0. –  Anonymous Sep 7 '11 at 15:42
@Anonymous: You probably know how to produce an open set $U_n$ containing the rationals having measure $\lambda(U_n) \leq 1/n$. Intersect. Get a dense $G_\delta$ of zero measure. See e.g. here (look at the comments). –  t.b. Sep 7 '11 at 16:06
@Theo: I must be very 'dense' today, but I still don't get it: Q is also a dense G_delta set of zero measure. But it's not exactly non-meager, is it? –  Anonymous Sep 7 '11 at 18:02
@Anonymous: Wait a minute. Yes, $\mathbb{Q}$ is dense, it is a meager $F_{\sigma}$ but it is not a $G_\delta$! –  t.b. Sep 7 '11 at 18:06
@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! –  Anonymous Sep 7 '11 at 18:16

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. –  Asaf Karagila May 13 '13 at 18:49

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