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?

**Answer**

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) \, .

$$

**Attribution***Source : Link , Question Author : Blathamani , Answer Author : Zen*