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.
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.