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I have been thinking about this question for a while now and found nothing on the matter so far.

Assuming the continuum hypothesis (or maybe also for the case that we assume that it is false), what is the cardinality of the set of measure zero subsets of the reals.

Obviously it is more than a continuum, but is it $\aleph_2$ or not?

Interested to hear your thoughts on the matter.

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    $\begingroup$ Any subset of a measure zero set is measure zero, so since there are measure zero sets of size $\mathfrak c$, there must be $2^{\mathfrak c}$ measure zero sets. So this doesn’t depend on cardinal arithmetic (though the value of $2^{\mathfrak c}$ does). $\endgroup$
    – spaceisdarkgreen
    Jun 7 at 12:31
  • $\begingroup$ Thats true, that's what I meant by it being larger than a continuum. But that's is only a lower bound, right? $\endgroup$
    – Staub und Dreck
    Jun 7 at 12:33
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    $\begingroup$ No, there are $2^{\mathfrak c}$ subsets of the reals, so it’s an upper bound as well. $\endgroup$
    – spaceisdarkgreen
    Jun 7 at 12:34
  • $\begingroup$ Oh my God you're right :) $\endgroup$
    – Staub und Dreck
    Jun 7 at 12:36

1 Answer 1

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The Cantor set is a set of measure zero, and there is a bijection between it and $\Bbb R$.

That means there is a bijection between their power sets. All subsets of the Cantor set have measure zero, so the cardinality of the set of all sets of measure zero is the same as the cardinality of the set of all subsets of $\Bbb R$.

The Continuum Hypothesis really doesn't matter for this question. $\#\Bbb R=2^{\aleph_0}=\mathfrak{c}$ and $\#\mathcal{P}(\Bbb R)=2^\mathfrak{c}$. If you want another notation for these cardinalities, the beth numbers provide that. By definition, $\beth_0=\aleph_0$, $\beth_1=2^{\beth_0}$, and $\beth_2=2^{\beth_1}$. So your set has cardinality $\beth_2$.

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  • $\begingroup$ Thanks you very much :) $\endgroup$
    – Staub und Dreck
    Jun 7 at 12:36

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