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A subset of $\mathbb{R}$ is said to be "negligible" iff it is of Lebesgue measure zero, and "meager" iff it is contained in a countable union of nowhere dense closed sets (i.e., closed sets with empty interior).

https://mathoverflow.net/questions/176488/can-we-define-an-empirically-generic-real-number claims to show a simple example of this, but I fail to see it in the question text. I also checked other questions here, but I didn't really see an explicit demonstration anywhere, nor a reference to one. So, how does one prove that claim?

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Desiato
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  • One easy way to break up the reals is into a set of rationals and irrationals. The rationals are a negligible set, so if (and I don't think it's possible) you can show that the irrationals are a meager set you are done. – graydad Oct 13 '14 at 03:35
  • The example given in the question partitions the reals into the set of oft-repeater in base $b$, which is negligible, and the complement of that set, which is meagre. (I’ve not actually checked the assertions; I’m just pointing out that the question does offer an example.) – Brian M. Scott Oct 13 '14 at 03:57
  • @bof that was my point – graydad Oct 13 '14 at 04:02

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Pick an enumeration of the rationals, say, $\mathbb Q=\{q_n:n\in\mathbb N\}$. Take $A_k=\bigcup_n (-\frac{1}{2^{n+k}}+q_n,q_n+\frac{1}{2^{n+k}})$. Now, consider $X=\bigcap_k A_k$, $X$ has measure zero because the measure of $\mu(A_k)\leq \frac{1}{2^k}$ and $X^c$ is meagre because is a countable union of closed nowhere dense sets (since each $A_k$ is dense open).

azarel
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