Cardinality of non-Borel sets

Question

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.

Answer 1

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.

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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".)

Answer 2

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.

Answer 3

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.