[proofplan] We prove discreteness directly from the defining countable-support criterion. Since $X$ is finite-valued, its values lie in a finite subset $S\subset\mathbb R$ with probability one, and every finite subset of $\mathbb R$ is countable. To also match the probability-mass-function formulation of discreteness, we enumerate $S$, assign masses to the atoms $\{X=x_i\}$, and verify that these masses sum to one. [/proofplan] [step:Enumerate the finite support and define its atom probabilities] Let $S\subset\mathbb R$ be a finite set satisfying $\mathbb P(X\in S)=1$. Since $\mathbb P(X\in\varnothing)=0$, the set $S$ is nonempty. Choose $n\in\mathbb N$ and distinct points $x_1,\dots,x_n\in\mathbb R$ such that \begin{align*} S=\{x_1,\dots,x_n\}. \end{align*} For each $i\in\{1,\dots,n\}$, define the event \begin{align*} A_i:=X^{-1}(\{x_i\})=\{\omega\in\Omega:X(\omega)=x_i\}. \end{align*} Because $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable and $\{x_i\}\in\mathcal B(\mathbb R)$, each $A_i$ belongs to $\mathcal F$. Define the mass assigned to $x_i$ by \begin{align*} p_i:=\mathbb P(A_i). \end{align*} The events $A_1,\dots,A_n$ are pairwise disjoint, and their union is the event $\{X\in S\}$. Therefore finite additivity of $\mathbb P$ gives \begin{align*} \sum_{i=1}^n p_i=\sum_{i=1}^n\mathbb P(A_i)=\mathbb P(X\in S)=1. \end{align*} [guided] The goal is to translate the statement "$X$ takes only finitely many values with probability one" into the standard data of a [discrete random variable](/page/Discrete%20Random%20Variable): a [countable set](/page/Countable%20Set) of possible values and probabilities attached to those values. Let $S\subset\mathbb R$ be a finite set such that $\mathbb P(X\in S)=1$. The empty set cannot have probability one, because $\mathbb P(X\in\varnothing)=0$, so $S$ is nonempty. Since $S$ is finite and nonempty, there are some $n\in\mathbb N$ and distinct [real numbers](/page/Real%20Numbers) $x_1,\dots,x_n\in\mathbb R$ such that \begin{align*} S=\{x_1,\dots,x_n\}. \end{align*} For each index $i\in\{1,\dots,n\}$, define the event \begin{align*} A_i:=X^{-1}(\{x_i\})=\{\omega\in\Omega:X(\omega)=x_i\}. \end{align*} This event is measurable: the singleton $\{x_i\}$ is a Borel subset of $\mathbb R$, and $X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable. Hence $A_i\in\mathcal F$, so its probability is defined. We set \begin{align*} p_i:=\mathbb P(A_i). \end{align*} Why are these the correct masses? The event $A_i$ is precisely the event that $X$ takes the value $x_i$. Since the points $x_1,\dots,x_n$ are distinct, the events $A_1,\dots,A_n$ are pairwise disjoint. Their union is exactly the event that $X$ lands somewhere in $S$: \begin{align*} \bigcup_{i=1}^n A_i=\{X\in S\}. \end{align*} By finite additivity of the probability measure $\mathbb P$ on pairwise disjoint events, \begin{align*} \sum_{i=1}^n p_i=\sum_{i=1}^n\mathbb P(A_i)=\mathbb P(X\in S)=1. \end{align*} Thus the finitely many values in $S$ account for all probability mass of $X$. [/guided] [/step] [step:Construct a probability mass function and conclude discreteness] Define the function \begin{align*} p_X:\mathbb R\to[0,1] \end{align*} as follows: for $x\in S$, set $p_X(x):=\mathbb P(X=x)$, and for $x\in\mathbb R\setminus S$, set $p_X(x):=0$. Since $S=\{x_1,\dots,x_n\}$ and $p_X(x_i)=p_i$ for each $i\in\{1,\dots,n\}$, we have \begin{align*} \sum_{x\in S}p_X(x)=\sum_{i=1}^n p_i=1. \end{align*} Also $S$ is finite, hence countable, and $\mathbb P(X\in S)=1$. Therefore $X$ takes values in the countable set $S$ with probability one. By the definition of a discrete real [random variable](/page/Random%20Variable), $X$ is discrete. [/step]