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<a_n\\          
  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.


*If X is a discrete r.v., then $P_x(C)=1$.
Thus,$F_X(x)=P(X\in[-\infty,x])=P(X\in[-\infty,x]\bigcap C)=\sum_{a_n\in[-\infty,x]}P_X({a_n})=\sum_{i=1}^\infty P_X({a_n})I\{a_n\leq x\}=\sum_{i=1}^\infty P_X(\{a_n\} )\delta_{a_n}(x)=\sum_{i=1}^\infty p_n\delta_{a_n}(x)$
 A: One direction is obvious: If $X$ is discrete, with probability distributed at points $a_1, a_2, \ldots$, then the atoms will be $\{a_1\}, \{a_2\}, \ldots$: clearly $P(\{a_i\}) = p_i$, and singletons with positive measure are always atoms. Now if $P(A) > 0$, and $a_{i_1}, a_{i_2}, \ldots$ are all $a_i$-s that belong to $A$, we have $P(A) = P(\{a_{i_1}, a_{i_2}, \ldots\}) = P(a_{i_1}) + P(a_{i_2}) + \ldots$, so the measure of every set is the sum of measures of all atoms contained in it, thus $P$ is atomic measure on $\mathbb{R}$.
Conversely, assume the probability distribution $P$ associated to random variable $X$ is atomic, meaning that $\mathbb{R}$ is a disjoint union $X_1 \cup X_2 \cup \ldots$, where every $X_i$ is either atom or null set (with respect to measure $P$). We'll show that every atom $X_i$ with positive measure can be written as $X_i' \cup \{a_i\}$, where $X_i'$ is null set. From this, the fact that $X$ is discrete, distributed at points $a_i$ easily follows.
Let $P$ be any measure on $\mathbb{R}$, and let $Y \subset \mathbb{R}$ be an atom with respect to measure $P$. Consider $Z_n = Y \cap [n, n+1)$ for $n \in \mathbb{Z}$. Since $Y$ is atom, exactly one of $Z_i$ has positive measure, let $Z_k$ be that one. Put $V_0 = Z_k$ and $U_0$ to be union of all remaining $Z_i$. We have $V_0 \cup U_0 = Y$, $V_0 \cap U_0 = \emptyset$, $P(U_0) = 0$. $P(V_0) = P(Y)$. Note that $V_0$ is now bounded in $\mathbb{R}$.
Since $V_0$ is fully contained in $[k, k+1)$, that is, interval of length 1, we can split this interval in half to obtain $A_1 = V_0 \cap [k, (k+1)/2), B_1 = V_0 \cap [(k+1)/2, k+1)$. Again, exactly one of those has positive measure, and we put it to be $V_1$, and we put $U_1$ to be union of $U_0$ and the one with null measure. Again we have $V_1 \cup U_1 = Y$, $V_1 \cap U_1 = \emptyset$, $P(U_1) = 0$. $P(V_1) = P(Y)$, and now $V_1$ is contained in an interval of length $1/2$.
We continue this procedure of splittint $V_i$ in half, setting $V_{i+1}$ to be the half with positive measure, and adding the other half to $U_i$ resulting with $U_{i+1}$. Finally, we set $U = \bigcup_{n = 0}^\infty U_i$, $V = \bigcap_{n = 0}^\infty V_i$. We have $U \cup V = Y$, $U \cap V = \emptyset$, $P(U) = 0$, $P(V) = P(Y)$, and since $V_i$ is contained in an interval of length $1/2^n$, $V$ must be a singleton, giving us the desired decomposition of $Y$ into union of a null set and singleton.
