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I was wondering concerning the following problem:

Take $X$ as a parameter space endowed with its Borel $\sigma$-algebra.
What is the cardinality of $\Delta (X)$, understood as the set of all probability measure over $X$?

[When I write that $X$ is a parameter space, I am thinking about the cardinality of $X$ as finite, countable or uncountable]

If $X$ is a doubleton, we should already have $| \Delta (X) | = \mathfrak{c}$.

What happens if we move on?
Also, how does this relate to the cardinality of the Borel sets?

Any feedback is most welcome.
Thank you for your time.

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    $\begingroup$ Is the $\sigma$-algebra fixed, or are we counting the couplets $(\mathcal E, P)$ such that $(X,\mathcal E,P)$ is a probability space? $\endgroup$ – user228113 Apr 15 '16 at 3:05
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    $\begingroup$ @Kolmin Then, can you say something about the $\sigma$-algebra you are considering? In quite a banal way, there is only one probability on $(X,\mathcal E)$ when $\mathcal E=\{\emptyset, X\}$. $\endgroup$ – user228113 Apr 15 '16 at 6:45
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    $\begingroup$ If $X$ is finite or countable, for a given $\sigma-$algebra the measure is defined by its value on singletons because probability is countably additive. In the countable case the cardinality is $\mathfrak c^{\aleph_0}=\mathfrak c$ $\endgroup$ – Ross Millikan Apr 20 '16 at 2:45
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    $\begingroup$ I don't now, that is why I didn't make it an answer. I suspect it gets larger, but depends on the specific $\sigma$-algebra. I think you can have a $\sigma$-algebra on $\mathfrak c$ where the sets are co-$\lt \mathfrak c$. Then you can choose any countable disjoint collection (in $\mathfrak c$ ways) to give some weight to, which gives at least $\mathfrak c^{\mathfrak c}$ measures. $\endgroup$ – Ross Millikan Apr 20 '16 at 4:12
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    $\begingroup$ : ) ... btw, what is the meaning of co-< $\mathfrak{c}$? $\endgroup$ – Kolmin Apr 20 '16 at 4:15
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This is only a partial answer (however, a complete question does not seem to be possible here).

Since you're mentioning Borel $\sigma$-algebra, I presume that $X$ is a topological space.

Extending Ross Millikan's comment, If $X$ has a countable base $B$, then $|\Delta(X)|=\mathfrak{c}$ unless the only non-empty open set is $X$. Indeed, in this case the measure is completely determined by its values on sets of the form $A_1\cap\dots\cap A_k$, $A_i\in B$ (since it is a $\pi$-system generating the Borel $\sigma$-algebra). But there are only countably many such sets. So the cardinality is at most $\mathfrak c$. On the other hand, it is at least $\mathfrak c$ by OP. In particular, for any separable metric space with at least two elements, $|\Delta(X)|=\mathfrak c$.

What can be said besides that, is an interesting question. Say, if $X$ is a non-separable metric space, then, obviously, $|\Delta(X)|\ge |X|$ (considering delta measures on singletons), and also $|\Delta(X)|\ge \mathfrak c$ (which would follow from the previous if we accept CH).

However, it is hard to bound $|\Delta(X)|$ from above. A helpful idea is that there cannot be an uncountable collection of disjoint sets of positive probability. In particular, the cardinality of set of all discrete probability measures is at most $(X\times \mathbb R)^{\aleph_0}$ (to each such measure we associate a sequence of points from $X$ and their corresponding weights), which is $\max(\mathfrak c, |X|)$.

Unfortunately, we cannot go much beyond this with such argument. We could write that each point from $X$ has a countable base of neighborhoods, and in total they make a base $B$ of topology. But the above argument will fail, since $B$ might not generate the whole Borel $\sigma$-algebra (since it is uncountable). It might still be that the values of this probability measure are completely determined by its values on some countable subfamily of $B$ (since, as I wrote, there can't be an uncountable disjoint family of sets with positive measure), but I am far from being sure in this.

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  • $\begingroup$ First of all, sorry for the late reply, and thanks for the answer. I have some questions: (1) Do you always assume $X$ uncountable? (2) Does your line of reasoning suggest, for example, that $| \Delta ([0, 1])| = \mathfrak{c}$? (3) what does OP mean in the body of the answer? (4) Could you clarifiy this "the measure is completely determined by its values on the base (since it is a $\pi$-system generating the $sigma$-algebra)"? $\endgroup$ – Kolmin Apr 21 '16 at 17:52
  • $\begingroup$ @Kolmin, yes, since $[0,1]$ is separable. $\endgroup$ – zhoraster Apr 21 '16 at 17:53
  • $\begingroup$ I was editing my comment. Now it is complete. $\endgroup$ – Kolmin Apr 21 '16 at 17:54
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    $\begingroup$ (1) no, why? (2) yes; (3) original post; (4) this is a corollary of Dynkin's theorem: if two probability measures agree on a $\pi$-system $\mathcal P$ (i.e. a family closed under intersection), then they coincide on the $\sigma$-algebra generated by $\mathcal P$. In our case this sigma-algebra contains all open sets, so it coincides with the Borel sigma-algebra. $\endgroup$ – zhoraster Apr 21 '16 at 18:19

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