Cardinality of a set such that $P(X=t)>0$

Question

$X$ is a random variable here, and $P$ is some probability. I have trouble understanding why would a set $$ \{ t\in \mathbb{R} : P(X = t)>0 \}, $$ have to be countable or finite.

Answer 1

For each $n\in\Bbb N$, consider the set $$ A_n=\{\,t\in\Bbb R:P(X=t)>\tfrac1n\,\}.$$ Then $$1\ge \sum_{t\in A_n}P(X=t)> \frac1n\left|A_n\right|$$ and hence $|A_n|$ is finite. Then $$ \bigcup_{n\in\Bbb N} A_n$$ is at most countably infinite.

Answer 2

Note that $\{t\in\mathbb R\mid P(X=t)>0\}=\bigcup_{n=1}^{\infty} A_n$ where $A_n=\{t\in\mathbb R\mid P(X=t)>\frac1n\}$.

So if $\{t\in\mathbb R\mid P(X=t)>0\}$ is not countable then some $n$ must exist such that $A_n$ is infinite.

This however leads to a contradiction.

If e.g. $B$ is an infinite and countable subset of this $A_n$ then $P(X\in B)=\sum_{b\in B}P(X=b)=+\infty$ which is absurd.

Answer 3

Consider $T_n=\{t \in \mathbb R: P(X=t) > \frac1{n} \}$.

The cardinality of each $T_n$ must be finite, because otherwise you would have $P(X \in T_n) = \infty$. Since $T = \cup T_n$, you get the result.