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Assume ZFC. Let $B\subseteq\mathbb R$ be a set that is not Borel-measurable. Clearly, $B$ must be uncountable, since countable sets are always Borel being a countable union of measurable singletons.

Question: can one conclude that $B$ necessarily has the cardinality of the continuum without assuming either the continuum hypothesis or the negation thereof?

A possibly related result is that any $\sigma$-algebra that contains infinitely many sets must necessarily have at least the cardinality of the continuum. This result is independent of the continuum hypothesis.

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No, you can't make those inferences. The Borel sets satisfy the continuum hypothesis. Namely any uncountable Borel set has necessarily the cardinality of the continuum.

If $B$ is not Borel measurable, in particular it is not a Borel set. All you can conclude about it is that it is uncountable. If the continuum is large, and $B$ is any cardinality between $\aleph_0$ and $2^{\aleph_0}$ then it is not Borel; and on the other hand there are plenty of non-Borel measurable (and not even Lebesgue measurable sets) of size continuum, for example Vitali sets.

In fact even if you replace Borel by Lebesgue you can't say much more. Martin's axiom implies that any set of size $<2^{\aleph_0}$ has Lebesgue measure zero, and in particular each non-measurable set must have size continuum. On the other hand adding an uncountable number of Random reals to any model of $\sf ZFC+GCH$ will add a non-Lebesgue measurable set of size $\aleph_1$, while possibly blowing up the continuum to be much larger. (Note that Random reals are a technical set theoretic term, not just "arbitrary reals chosen at random".)

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  • $\begingroup$ Thank you for your answer. I think I understand now that the question does not have a definitive answer without further information on $B$ if the negation of CH is assumed. FYI, the downvote you got wasn't from me. $\endgroup$
    – triple_sec
    Nov 11, 2014 at 22:58
  • $\begingroup$ You're welcome! $\endgroup$
    – Asaf Karagila ♦
    Nov 12, 2014 at 3:40
  • $\begingroup$ In the last sentence, you probably meant random (not Cohen) reals. $\endgroup$
    – hot_queen
    Nov 16, 2014 at 21:08
  • $\begingroup$ @hot_queen: Not according to Blass' chapter in the Handbook I didn't. Or maybe I could have meant Random as well. $\endgroup$
    – Asaf Karagila ♦
    Nov 16, 2014 at 21:10
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    $\begingroup$ Sorry, just one more comment and I'll stop: You don't need GCH in V. The random reals that you added constitute a Sierpinski set (an uncountable set of reals all of whose uncountable subsets are non null). The reason is that any null Borel set appearing in the extension can be decoded with countably many random reals so the other random reals avoid this null set. $\endgroup$
    – hot_queen
    Nov 16, 2014 at 21:37
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The situation is quite the opposite. Every Borel set is either countable or has size continuum. So if $B \subseteq \mathbb{R}$ is an uncountable set of size less than the continuum, it is not Borel-measurable.

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  • $\begingroup$ The original poster intended for $B$ to be a non-Borel set, not a Borel set, and the question is what can be concluded about the cardinality of a non-Borel set (besides the obvious fact that it must be greater than $\aleph_{0}$ and less than or equal to $2^{\aleph_{0}}).$ $\endgroup$
    – Dave L. Renfro
    Nov 11, 2014 at 19:56
  • $\begingroup$ @DaveL.Renfro Actually, I think this almost answers my question, which apparently hinges crucially on CH. Indeed, if CH is assumed, then every non-Borel set has necessarily the cardinality of the continuum. If, on the other hand, the negation of CH is assumed, then any $B$ whose cardinality is strictly between $\#\mathbb N$ and $\#\mathbb R$ (and such sets do exist given $\lnot$CH) is necessarily non-Borel. Moreover, there exist non-Borel sets whose cardinalities are $\#\mathbb R$ (e.g., Vitali set), so we cannot conclude anything a priori. $\endgroup$
    – triple_sec
    Nov 11, 2014 at 22:51
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You can still have an idea of the cardinality of sets that are definable in a more general sense than being Borel. Say a set is analytic $(\bf{\Sigma}^1_1)$ if it is the continuous image of a Borel set, coanalytic $(\bf{\Pi^1_1})$ if its complement is analytic, and $\bf{\Sigma}^1_2$ if it is the continuous image of a coanalytic set.

As is the case with Borel sets, analytic sets have the perfect set property: either they are countable or they contain a perfect subset. Since perfect sets have size $\mathfrak{c}$, analytic sets satisfy the continuum hypothesis.

The result for $\bf{\Sigma}^1_2$ sets (which include coanalytic sets) is close, but not quite CH. These sets can have cardinality $\aleph_0, \aleph_1$ or $2^{\aleph_0}$, for the reason that they can be expressed as the union of $\aleph_1$ Borel sets.

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