Probability mass function, a proof of the basic property.

Question

Lemma

The probability mass function $f:\mathbb{R}\to[0,1]$ satisfies, the set of $x$ such that$f(x)\ne0$ is countable.

Could you help me to prove it? The textbook does not provide any proof or hint. It only says that the proof is obvious.

Answer 1

Because of $$A_n:=\{x: f(x)>1/n\} \implies \{x: f(x)\neq 0\}=\bigcup_{n\in\mathbb{N}} A_n $$

it is enough to show that the $A_n$ are finite. Since countable unions of countable sets are countable.

Now assume that $A_n$ is infinite. So I can find a sequence $x_i \subset A_n$ of distinct points. Thus: $$P(A_n)\ge P\left(\biguplus_{i\in\mathbb{N}} \{x_i\}\right) = \sum_{i=1}^\infty P(\{x_i\})= \sum_{i=1}^\infty f(x_i)>\sum_{i=1}^\infty 1/n = \infty$$

Which is a contradiction.

In words: The number events with probability above $1/n$ is finite. By placing these sets of events after each other it becomes obvious that I can count through all events which have positive probability.