- #1

psie

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- 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

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.

ClaimGiven 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.