I The set of nonzero values of pmf is at most countable

[psie]

TL;DR Summary

As the title implies, I'm working on showing this relatively basic exercise. I was wondering if my solution is correct. The crux of my argument is that every infinite set has a countable infinite subset.

The claim is

Claim Given a discrete random variable $X$ taking values in $\mathbb R$, the number of $x$ values such that $p_X(x)=P(X=x)>0$ is at most countable.

Proof. If it was $P(X = x ) > 0$ for uncountably many $x$, then there would exist an $N\in\mathbb N$ such that $P(X = x) > \frac{1}{N}$ for uncountably many $x$ (1). This in turn would imply that there exist countably infinite many $x$ such that $P(X=x)>\frac{1}{N}$, since every infinite set has a countably infinite subset. Denote this subset by $A=\{x_1,x_2,\ldots\}$. Then $$P(X\in A) = \sum_{i\in \mathbb N} P(X = x_i) >\sum_{i\in \mathbb N}\frac1{N}=\infty,$$contradicting $P(X\in A)\leq 1$.

(1) if such an $N$ would not exist, then $A_n=\{x\in\mathbb R: P(X = x) > \frac{1}{n}\}$ would be countable for every $n$, and the countable union of countable sets is countable, i.e. $\{x\in\mathbb R: P(X=x)>0\}=\bigcup_{n\in\mathbb N}A_n$ is countable.

Thoughts, comments? I guess a more general claim is that any finite measure has at most countably many atoms.

 

[psie]

Writing this down, I realize one could simply argue as follows too (without contradiction). Since the measure space is finite, the number of singletons with measure $1/n$ or more is finite. Any positive number here will do, because their total mass will be at least their number times that lower bound, and since the space is finite, their number has to be finite. Now, for each $n$ there are finitely many singletons of measure greater than $1/n$, and the set of all singletons of positive measure is the union of those. And a countable union of finite sets is countable.

 

[nuuskur]

Here's how I'd formulate it. Let $C$ be a set of positive numbers. Define $C_n = \{x\in C \mid x> 1/n\}$. Then $C_n$ is a monotone increasing sequence w.r.t inclusion and $C = \bigcup C_n$.

Note that for each $n\in\mathbb N$ we have
<br /> \sum _{x\in C} x \geqslant \sum _{x\in C_n} x \geqslant \frac{|C_n|}{n}.<br />
But the sum of the probabilities of atoms is bounded (by $1$), hence the sets $C_n$ must be finite, therefore $C$ can be at most countable.

 

[WWGD]

Edit: A sort of combination of your two arguments:
Yes, the only way an uncountable sum will converge is if it has finite support. You can partition your support by $S_n:= x, x\geq 1/n$ . Then, for an infinite sum, at least one of the $S_n$ will be infinite, so you have an infinite number of copies of $1/n$ as a lower bound of your sum, and this will diverge.