measure theory - How to proof every r.v. in a probability space is discrete if and only if the pm. is atomic?

Definition 1: As for the definition atomic measure,
Given a measurable space $(X,\sum)$ and a measure $\mu$ on that space, a set $A\subset X$ in $\Sigma$ is called an atom if and for any measurable subset
$B\subset A$ with $\mu (B)<\mu (A)$ the set $B$ has measure zero. Thus B either has measure $\mu(A)$ or has measure zero.

Definition 2:According to the definition of discrete random varaible:

A random variable $X$ on $(\Omega, A, P)$ is discrete if $\exists$ a countable subset $C$ of $R$, s.t. $P(X\in C)=1$ (According to the defintion of A course in probability theory, Kai Lai Chung)

I am not sure about the proof.

Though there exists some relationship between the probability measure and the random variable if they are discrete.

A d.f F is called discrete if it can be represented in the form
$F(x) =\sum_{n=1}^{+\infty} p_n\delta_{a_n}(x)$
with $\delta_{a_n}(x)$ is degenerate such that

\begin{equation}

\delta_{a_n}(x)=
\left\{
\begin{array}{lr}
0, & x 1, & x\geq a_n.\\
\end{array}\right.\end{equation}

1.We can show put $C=\{a1,a2,a_n\dots\}$if $F_X$ is discrete, then $F(x) =\sum_{n=1}^{+\infty} p_n\delta_{a_n}(x)$ where $\sum_{n=1}^{+\infty} p_n=1$

Thus $P_X(C)=P_X(\bigcup_{n=1}^{+\infty}\{a_n\} )=\sum_{n=1}^{+\infty}P_X(\{a_n\} )=\sum_{n=1}^{+\infty}[F_X(a_n)-F_X(a_n^-)]=\sum_{n=1}^{+\infty}p_n=1$

Thus, X is discrete r.v.

2. If X is a discrete r.v., then $P_x(C)=1$.

   
   
   

   Thus,$F_X(x)=P(X\in[-\infty,x])=P(X\in[-\infty,x]\bigcap C)=\sum_{a_n\in[-\infty,x]}P_X({a_n})=\sum_{i=1}^\infty P_X({a_n})I\{a_n\leq x\}=\sum_{i=1}^\infty P_X(\{a_n\} )\delta_{a_n}(x)=\sum_{i=1}^\infty p_n\delta_{a_n}(x)$