# How can I prove that the cumulative distribution function is right continuous?

I've learned in my probability courses that the cumulative distribution function $F$ of a random variable $X$ is right continuous. Is it possible to prove that?

To prove the right continuity of the distribution function you have to use the continuity from above of $$P$$, which you probably proved in one of your probability courses.

Lemma. If a sequence of events $$\{A_n\}_{n\geq 1}$$ is decreasing, in the sense that $$A_n\supset A_{n+1}$$ for every $$n\geq 1$$, then $$P(A_n)\downarrow P(A)$$, in which $$A=\cap_{n=1}^\infty A_n$$.

Let's use the Lemma. The distribution function $$F$$ is right continuous at some point $$a$$ if and only if for every decreasing sequence of real numbers $$\{x_n\}_{n\geq 1}$$ such that $$x_n\downarrow a$$ we have $$F(x_n)\downarrow F(a)$$.

Define the events $$A_n=\{\omega : X(\omega)\leq x_n\}$$, for $$n\geq 1$$. We will prove that $$\bigcap_{n=1}^\infty A_n=\{\omega:X(\omega)\leq a\}\, .$$

In one direction, if $$X(\omega)\leq x_n$$ for every $$n\geq 1$$, since $$x_n\downarrow a$$, we have $$X(\omega)\leq a$$.

In the other direction, if $$X(\omega)\leq a$$, since $$a\leq x_n$$ for each $$n\geq 1$$, we have $$X(\omega)\leq x_n$$, for every $$n\geq 1$$.

Using the Lemma, the result follows: $$F(x_n) = P\{X\leq x_n\} = P(A_n) \downarrow P\left( \cap_{n=1}^\infty A_n \right) = P\{X\leq a\} = F(a) \, .$$