# A question about nonmeasurable subset of real line and uncountable measure-zero subset. [closed]

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.

• Does "measurable" here mean Lebesgue measurable or Borel measurable? – Nate Eldredge Apr 6 at 15:58
• @NateEldredge Lebesgue measure. – algebra.And.analysis Apr 6 at 16:18
• (For other readers: here is a proof of the mentioned result for measurable sets.) – Noah Schweber Apr 6 at 17:02
• Please edit your question to include an actual question in some place other than the title. – Xander Henderson Apr 7 at 2:26
• @XanderHenderson I havr edited it. – algebra.And.analysis Apr 7 at 3:30

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.

• Honestly,I am inclined to think that CH(or GCH) is right.Set theory is very complicated..... – algebra.And.analysis Apr 6 at 18:05
• Indeed, in the second bullet we can take $\mathcal{G}$ to be the set of all measure-zero $G_\delta$ sets. – Nate Eldredge Apr 6 at 18:23
• @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. – Noah Schweber Apr 6 at 19:07
• 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). – Asaf Karagila Apr 7 at 6:59