Is the set of probability distributions infinite in measure? I've been learning about more probability distributions that are not covered in introductory statistics, and I realized looking at Wikipedia that there are a lot that have been named.
This got me wondering, is there a countably infinite number of probability distributions?
 A: We'll show that the set of probability distributions over the reals has cardinality $2^{\aleph_0},$ the same as the cardinality of $\mathbb{R}.$
The cardinality of the set $S$ of probability distributions is the same as the cardinality of the set $T$ of cumulative distribution functions, since there's a clear one-to-one correspondence between these sets.
A cumulative distribution function is a function $f\colon \mathbb{R}\to\mathbb{R}$ such that the following four conditions hold:
(a) $\;f$ is monotonically increasing, that is, $x\le y\implies f(x)\le f(y),$
(b) $\;\lim_{x\to-\infty}f(x)=0,$
(c) $\;\lim_{x\to\infty}f(x)=1,$
(d) $\;f$ is right-continuous.
Let $M$ be the set of monotonic functions from $\mathbb{R}$ to $\mathbb{R}.$  We'll show that
$$2^{\aleph_0} \le \operatorname{card}(T) \le  \operatorname{card}(M) \le 2^{\aleph_0}.$$
Since $S$ and $T$ have the same cardinality, that's sufficient to show that $S$ has the cardinality of the continuum.
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(1) $2^{\aleph_0} \le \operatorname{card}(T).$
Proof of (1): For each real number $r,$ the step function which maps any $x\lt r$ to $0$ and any $x \ge r$ to $1$ is a cumulative probability function, and these step functions are all distinct (so the map which takes $r$ to the corresponding step function is one-to-one).
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(2) $\operatorname{card}(T) \le \operatorname{card}(M).$
Proof of (2): $T\subseteq M.$
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(3) Any monotonic function from $\mathbb{R}$ to $\mathbb{R}$ has only countably many points of discontinuity.
Proof of (3):
Let $f$ be monotonic; without loss of generality, we'll assume that $f$ is monotonically increasing. Let $D$ be the set of points at which $f$ is discontinuous.  For each $d\in D,$ $\lim_{x\to d^-}\lt \lim_{x\to d^+},$ so there is a rational number $q$ such that $\lim_{x\to d^-}\lt q\lt\lim_{x\to d^+}.$ Using some standard enumeration of $\mathbb{Q},$ let $j$ be the function which maps each $d\in D$ to the first rational number $q$ in the enumeration with the property that $\lim_{x\to d^-}\lt q\lt\lim_{x\to d^+}.$ 
We can see that $j$ is one-to-one on $D$ as follows: If $d_1\lt d_2$ are both in $D,$ then, by monotonicity of $f,$ any $x$ in the open interval $(d_1,d_2)$ satisfies $f(d_1)\le f(x) \le f(d_2).$  It follows that $\lim_{x\to d_1^{\,+}}f(x) \le \lim_{x\to d_2^{\,-}}f(x),$ so we have $j(d_1)\lt \lim_{x\to d_1^{\,+}}f(x) \le \lim_{x\to d_2^{\,-}}f(x) \lt j(d_2).$
We now know that $j$ maps $D$ one-to-one into $\mathbb{Q};$ $\mathbb{Q}$ is countable, so $D$ is countable, completing the proof of (3).
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(4) $\operatorname{card}(M) \le 2^{\aleph_0}.$
Proof of (4):
Let $W$ be the set of all countable subsets of $\mathbb{R}\times\mathbb{R}.$  Define a function $K\colon M\to W$ by setting $K(f)=\lbrace \langle x,f(x)\rangle | x\in\mathbb{Q}\text{ or }f\text{ is discontinuous at }x\rbrace.$
For any $f\in M,$ $K(f)$ belongs to $W,$ by (3). 
Since $W$ has cardinality $2^{\aleph_0},$ we just need to show that $K$ is one-to-one to conclude that $M$ also has cardinality less than or equal to $2^{\aleph_0}.$  But if $K(f)=K(g),$ then $f$ and $g$ are equal on the set $\mathbb{Q},$ which is dense in $\mathbb{R}.$  It follows that $f(x)=g(x)$ for every $x\in\mathbb{R}$ at which $f$ and $g$ are both continuous.  On the other hand, if $f$ or $g$ is discontinuous at $x,$ then some ordered pair in  $K(f)=K(g)$ has $x$ as its first coordinate.  If $y$ is the second coordinate of this ordered pair, then $f(x)=y=g(x).$  So $f(x)=g(x)$ for all $x\in\mathbb{R}.$ This completes the proof that $K$ is one-to-one and hence that $M$ has cardinality at most $2^{\aleph_0}.$
