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Having recently studied a course on $\mathsf{ZFC}$ Set Theory, I have been thinking about applications of the material that I have learnt in other disciplines of mathematics. With that in mind, I recently formulated the following question and have been unable to find anything definitive online:

Question: What is the cardinality of the set ${A}$ containing all measures $\mu$ on $(\mathbb{R}, \mathcal{B} (\mathbb{R}))$? (where $\mathcal{B} (\mathbb{R} )$ refers to the Borel Sigma Algebra on $\mathbb R$)

I know that the cardinality of $\mathcal{B}(\mathbb{R})$ is $2^{\aleph _0}$ (with a proof of this result here) and it feels as though we should be able to construct a suitable upper bound using this fact. We know that a measure on this space will be a function $\mu : \mathcal{B} (\mathbb{R}) \rightarrow [0, + \infty ]$. Therefore, the cardinality of the set of measures is certainly upper bounded by the cardinality of $[0, + \infty ]^{\mathcal{B}(\mathbb{R})} = \mathbb{R}^{\mathbb{R}}$. This is $\big{(}2^{\aleph _0} \big{)}^{2^{\aleph _0}}$. So we have: $$|A| \leq \big{(}2^{\aleph _0} \big{)}^{2^{\aleph _0}} = 2^{2^{\aleph _0}}$$

For a lower bound, we note that each probability distribution on $\mathbb{R}$ induces its own measure. And the set of the probability distributions on $\mathbb{R}$ has cardinality $2^{\aleph _0}$ (with a proof of this result here). So we have: $$2^{\aleph _0} \leq |A| $$

We now have an upper bound and a lower bound for the cardinality of the set $A$ in question. This allows us to produce the following inequality (by combining the two that we have above):

$$ 2^{\aleph_0} \leq |A| \leq |\mathbb{R}^{\mathbb{R}}| = 2^{2^{\aleph _0}}$$

Perhaps it is easier to first consider the the cardinality of the set of probability measures first in order to simplify the problem, before considering all measures - although I haven't managed to have much success in tightening the bounds this way.

But this inequality still leaves me wondering what the exact cardinality is from a rigorous perspective. Is there a way to establish tighter bounds to find the probability exactly and is what I have done so far correct?

I would be grateful for any assistance in either tightening the bounds or finding the exact cardinality.

FD_bfa
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1 Answers1

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There are $2^{2^{\aleph_0}}$ Borel measures on $\mathbb{R}$. Indeed, for any subset $A\subseteq\mathbb{R}$, there is counting measure on $A$, where $\mu(X)=|X\cap A|$ if $X\cap A$ is finite and $\mu(X)=\infty$ otherwise.

On the other hand, there are only $2^{\aleph_0}$ Borel probability measures on $\mathbb{R}$. This follows, for instance, from the fact that such a measure is always outer regular, so it is determined by its values on open sets. The values on open sets are in turn determined by the values on open intervals with rational endpoints, since every open set is the union of an increasing sequence of disjoint unions of open intervals with rational endpoints. There are only countably many such intervals, so there are only $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$ different values a measure can take on them.

Eric Wofsey
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