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Does every nonmeasurable subset of real line have an uncountable and measure-zero subset?

Just consider the Lebesgue measure.

I have known that this is true for every measurable set whose measure is greater than zero.

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    $\begingroup$ Does "measurable" here mean Lebesgue measurable or Borel measurable? $\endgroup$ – Nate Eldredge Apr 6 at 15:58
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    $\begingroup$ @NateEldredge Lebesgue measure. $\endgroup$ – algebra.And.analysis Apr 6 at 16:18
  • $\begingroup$ (For other readers: here is a proof of the mentioned result for measurable sets.) $\endgroup$ – Noah Schweber Apr 6 at 17:02
  • $\begingroup$ Please edit your question to include an actual question in some place other than the title. $\endgroup$ – Xander Henderson Apr 7 at 2:26
  • $\begingroup$ @XanderHenderson I havr edited it. $\endgroup$ – algebra.And.analysis Apr 7 at 3:30
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This cannot be answered by the usual axioms of set theory.

  • It's consistent with ZFC that every set of reals of size $\aleph_1$ has measure zero (note that this requires $2^{\aleph_0}>\aleph_1$). For example, this follows from MA$_{\aleph_1}$. In this case if $A$ is non-measurable, just consider any subset of $A$ of cardinality $\aleph_1$.

  • Meanwhile, a consistent negative answer is easy to construct via transfinite recursion assuming CH (these are the Sierpinski sets). The key point here is that, while there are $2^{2^{\aleph_0}}$-many null sets in general, we can find a set $\mathcal{G}$ of $2^{\aleph_0}$-many (hence by CH, $\aleph_1$-many) null sets such that every null set is contained in one in $\mathcal{G}$, and this lets us set up a length-$\omega_1$ recursion (length-$\omega_1$ is useful since it means that at each step we've only "used countably many points" so there are lots of points "still untouched" as we continue to build our set).

Finally, I don't know what happens if both CH and MA$_{\aleph_1}$ fail.


As an aside, note the crucial role in the first bulletpoint of the number $$\mathfrak{m}=\mbox{the least size of a non-measurable set}.$$ This "$2^{\aleph_0}$-like" number is a cardinal characteristic of the continuum - there's a lot of interesting mathematics around what we can (and can't!) prove about CCCs.

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    $\begingroup$ Honestly,I am inclined to think that CH(or GCH) is right.Set theory is very complicated..... $\endgroup$ – algebra.And.analysis Apr 6 at 18:05
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    $\begingroup$ Indeed, in the second bullet we can take $\mathcal{G}$ to be the set of all measure-zero $G_\delta$ sets. $\endgroup$ – Nate Eldredge Apr 6 at 18:23
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    $\begingroup$ @algebra.And.analysis FWIW I tend to prefer $\neg CH$ - it makes the theory of cardinal characteristics nontrivial, and is compatible with various other fun shenanigans. But I go back and forth. $\endgroup$ – Noah Schweber Apr 6 at 19:07
  • $\begingroup$ Adding $\aleph_1$ Cohen reals will always add a Luzin set. Or maybe Random reals. One of the two, I never remember which is which (the other adds a similar set for Baire). $\endgroup$ – Asaf Karagila Apr 7 at 6:59

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