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$\aleph_1$ is the cardinality of the countable ordinals. It is the least cardinal number greater than $\aleph_0$, and assuming the continuum hypothesis it's equal to $\mathfrak{c}$, the cardinality of the the real numbers.

My question is, is it possible for all $\aleph_1$ subsets of $\Bbb{R}$ to have Lebesgue measure $0$? This is, of course, impossible assuming the continuum hypothesis, because then all sets of real numbers would have measure $0$, which is absurd. But is $\mathsf{ZFC}$ + $\lnot\mathsf{CH}$ + "all subset of $\Bbb{R}$ of cardinality $\aleph_1$ have Lebesgue measure $0$" consistent? If not, what if we replaced Lebesgue measure with some other measure?

It may be worth noting that, although there may be subsets of $\Bbb{R}$ with cardinality $\aleph_1$, there are no subsets of $\Bbb{R}$ which have the order-type $\omega_1$ (the order-type of the countable ordinals) under the usual ordering on $\Bbb{R}$.

Any help would be greatly appreciated.

Thank you in Advance.

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are you talking about an outer measure, because it is not clear that $\aleph_1$ sets need to be measurable at all –  Dominic Michaelis Oct 21 '13 at 12:48
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If Martin's axiom (and negation of CH) holds, then union of $\aleph_1$ sets of Lebesgue measure 0 also Lebesgue measure 0. In particular, a set has cardinality $\aleph_1$ has Lebesgue measure 0. It is known that if ZFC is consistent, then $\mathsf{ZFC+\lnot CH+MA}$ also consistent. –  tetori Oct 21 '13 at 12:53
    
@DominicMichaelis A set which has outer measure zero is always measurable, so "measure zero" and "outer measure zero" are synonymous. –  Keshav Srinivasan Oct 21 '13 at 14:10
    
@tetori Thanks. If you post it as an answer, I'm happy to accept it. –  Keshav Srinivasan Oct 21 '13 at 14:11
    
@KeshavSrinivasan you asked for other measures too –  Dominic Michaelis Oct 21 '13 at 19:11

3 Answers 3

Of course assuming CH they can. But if $\aleph_1 < 2^{\aleph_0}$ then there is no measurable set of cardinality $\aleph_1$ having positive Lebesgue measure, so the answer to the question in the title is no. We can prove the following in ZFC without requiring additional axioms.

Proposition (ZFC). Every uncountable Borel subset of $\mathbb{R}$ has cardinality $2^{\aleph_0}$.

This is a corollary of the Perfect Set Theorem (every uncountable Borel set contains a perfect set, i.e. a set with no isolated points, and any such set has cardinality at least $2^{\aleph_0}$). For a proof, see Theorem 13.6 of Kechris's Classical Descriptive Set Theory.

Corollary (ZFC). Every Lebesgue measurable subset of $\mathbb{R}$ with cardinality less than $2^{\aleph_0}$ has Lebesgue measure zero.

Proof. Every Lebesgue measurable set $E$ can be written $E = B \cup N$ where $B$ is Borel and $m(N) = 0$; in particular $m(E) = m(B)$. But if $|E| < 2^{\aleph_0}$ then $|B| < 2^{\aleph_0}$, so by the previous proposition $B$ is countable and hence $m(B) = 0$.

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This, however, does not mean that every set of reals of size $\aleph_1$ is indeed Lebesgue measurable. –  Asaf Karagila Oct 21 '13 at 15:09
    
@AsafKaragila: True. I guess I answered a slightly different question. –  Nate Eldredge Oct 21 '13 at 15:36
    
@NateEldredge (Thanks for the edit.) Since people have mentioned $\mathsf{MA}$, let me add that under that axiom, every set of size below the continuum is measurable. In terms of cardinals characteristics of the continuum, the assumption relevant to the question (and implied by $\mathsf{MA}+\lnot\mathsf{CH}$) is denoted by $\mathrm{non}(\mathcal L)>\aleph_1$. –  Andres Caicedo Oct 22 '13 at 0:28

A very simple model of "$2^{\omega} > \omega_1$ and every set of reals of size $\omega_1$ is null" is the Cohen's original model for the failure of CH. It is obtained by adding $\omega_2$ Cohen reals - i.e. forcing with $Fn(\omega_2, 2)$. The proof uses the fact that adding a Cohen real makes the set of old reals null. The category analogue of this can be obtained by adding $\omega_2$ random reals - i.e. forcing with the usual measure algebra on $2^{\omega_2}$.

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I actually had a hunch that this is true. But the MA argument has a slightly "cleaner" proof. –  Asaf Karagila Oct 21 '13 at 20:02
    
Both models are fine. This one just happens to avoid iterated forcing. –  hot_queen Oct 21 '13 at 20:08
    
Yes, but it's easier to take the consistency of $\sf MA+\lnot CH$ as a blackbox, in which case the argument becomes simpler. –  Asaf Karagila Oct 21 '13 at 20:09
    
@AsafKaragila I disagree. $\mathsf{MA}$ identifies all sorts of invariants, so it is too strong and full of distractions for the task at hand, and the black-box approach hides the actual reason why a result holds. (I agree that it helps if one has no interest in the actual combinatorics behind the consistency of the statement.) –  Andres Caicedo Oct 22 '13 at 0:32
    
@Andres, the way I read the question fits the remark in parenthesis; so you do agree with me. :-) –  Asaf Karagila Oct 22 '13 at 4:30

The following result is due to Martin and Solovay, and can be found in Jech's Set Theory (3rd Millennium edition) as theorem 26.39.

If Martin's Axiom holds, then the union of fewer than $2^{\aleph_0}$ null sets is null, and the union of fewer than $2^{\aleph_0}$ meager sets is meager.

It follows, if so, that under $\sf ZFC+MA+\lnot CH$ the union of $\aleph_1$ singletons has measure zero, so every set of size $\aleph_1$ has measure zero.

It suffices, if so, to show that $\sf ZFC+MA+\lnot CH$ is a consistent theory (relative to the consistency of $\sf ZFC$, of course). This is a result of Solovay and Tennenbaum, which appears in the same book as Theorem 16.13.

Assume $\sf GCH$ and let $\kappa$ be a regular cardinal greater than $\aleph_1$. There is a c.c.c. notion of forcing $P$ such that the generic extension $V[G]$ by $P$ satisfies Martin's Axiom and $2^{\aleph_0}=\kappa$.

By taking, for example, $V=L$ as the ground model and $\kappa=\aleph_2$, the result is a model where the continuum is $\aleph_2$ and the union of $\aleph_1$ null sets is a null set.

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