# 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

$$$$\delta_{a_n}(x)= \left\{ \begin{array}{lr} 0, & x

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.

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

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

• Your definition of "atomic measure" is incomplete. In the definition of "atom," I think you mean $$\text{[...] } A \subset X \text{ in } \Sigma \text{ is called an atom if } \color{red}{\mu(A) > 0} \text{ and for any [...]}$$ Then you're missing what "atomic measure" means. Perhaps something like "$X$ is the disjoint union of atoms"? – Brian Moehring Oct 8 at 20:53
• What is "pm" in the title (it is better not to use abbreviation)? Does it mean probability measure induced by $X$? If so, you may also need to explain what is meant by "a probability measure is atomic". – Danny Pak-Keung Chan Oct 8 at 21:30

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.