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I have learnt from this question that comeagre null sets (sets with Lebesgue measure zero whose complement is a countable union of nowhere dense sets) do exist, and they cannot be included in $F_\sigma$ null sets. It seems that being comeagre means that a set is "big", and being null means that a set is "small". I would like to learn more about such kind of sets.

So my questions are:

(a) Does there exist a comeagre null set which is non-Borel?

(b) Is it true that: if $E$ is a comeagre null set that is non-Borel, $B$ is a Borel subset of $E$, then $E\setminus B$ does not have a superset which is a $F_\sigma$ null set?

Any help appreciated.

EDIT: As stated in the comment, there exists a comeagre null set by a cardinality argument. It is standard that there are $\mathfrak{c}$ Borel sets of $\mathbb{R}$. Pick any comeagre null set $E_0$, let $C$ be the standard Cantor set, then $E_0\setminus C$ is a comeagre null set disjoint with $C$, and for any set $S\subset C$ we have $(E_0\setminus C)\cup S$ being comeagre null. So there are $2^\mathfrak{c}$ comeagre null sets. This shows that (a) is true.

If we take $E_0$ to be Borel (take the standard construction for example), $S\subset C$ to be non-Borel, then $E = (E_0\setminus C)\cup S$ is non-Borel, but $C$ is a $F_\sigma$ null set that is a superset of $E \setminus (E_0\setminus C) = S$. This shows that (b) is false.

Jianing Song
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  • If you don't get an answer straight away, this is probably hard enough for Math Overflow. – Chris Sanders May 03 '22 at 12:30
  • A great little book: Measure and Category by John C. Oxtoby. See if your library has it! – GEdgar May 03 '22 at 12:56
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    See if you can prove: there are $\mathfrak{c} = 2^{\aleph_0}$ Borel sets, but there are $2^{\mathfrak{c}}$ comeager null sets in $\mathbb R$. – GEdgar May 03 '22 at 12:59
  • @ChrisSanders Yes, but good answers take time, so I'll wait :) – Jianing Song May 03 '22 at 14:58
  • @GEdgar Yes, that's great hint. It is standard that there are $\mathfrak{c}$ Borel sets of $\mathbb{R}$. Pick any comeagre null set $E$, let $C$ be the standard Cantor set, then $E\setminus C$ is a comeagre null set disjoint with $C$, and for any set $S\subset C$ we have $(E\setminus C)\cup S$ being comeagre null. So there are $2^\mathfrak{c}$ comeagre null sets. Thanks! – Jianing Song May 03 '22 at 14:58
  • FYI, "$F_{\sigma}$ null" $\implies$ "meager and null" (simultaneously), but not conversely. See this 1 May 2000 sci.math post and my answer to Jordan measure zero discontinuities a necessary condition for integrability. Incidentally, while "$F_{\sigma}$ null" sets are small in the sense of measure and category, the 3rd paragraph of this answer (continued) – Dave L. Renfro May 03 '22 at 18:07
  • shows that such a set can have Hausdorff dimension $1$ in every nonempty open interval (a much stronger "largeness condition" than continuum many points in every nonempty open interval). Note: In my previous comment, "not conversely" of course is obvious because one simply picks a "meager and null" set that is not $F_{\sigma}.$ What I meant is that there exists a "meager and null" set that is not the subset of any "$F_{\sigma}$ null" set (or even any countable union of "$F_{\sigma}$ null" sets, although this additional strengthening is automatic). – Dave L. Renfro May 03 '22 at 18:17
  • @DaveL.Renfro Thanks for your information! – Jianing Song May 04 '22 at 03:35

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