> [!NOTE] Proposition > Let $f:\mathbb{R}\to[0,1]$ be a [[Probability Mass Function|probability mass function]]. Then there exists a [[Probability Space|probability space]] $(\Omega,\mathcal{F},\mathbb{P})$ and a [[Discrete random variables|discrete real-valued random variable]] $X$ such that $p_{X}(x)=f(x)$ for every $x\in \mathbb{R},$ where $p_{X}$ denotes the [[Probability Mass Function of Discrete Real-Valued Random Variable|probability mass function]] of $X.$ **Proof**: Let $D=\{ x:f(x)>0 \}.$ Then $D$ is [[Countable Set|countable]] since $f$ is a probability mass function. Take $\Omega = \mathbb{R},\mathcal{F}=\mathcal{B}(\mathbb{R})$ and $X:\Omega\to \mathbb{R}$ defined by $X(x)=x$ for all $x\in \Omega=\mathbb{R}.$ Define $\mathbb{P}:\mathcal{B}(\mathbb{R})\to \mathbb{R}$ by $\mathbb{P}(B)= \sum_{x\in D \cap B} f(x).$Then $\mathbb{P}$ is a [[Probability Measure|probability measure]] on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ since it satisfies Kolmogorov's axioms: (1) Let $B\in \mathcal{B}(\mathbb{R}).$ Then $B \cap D \subset D.$ Thus $\mathbb{P}(B)= \sum_{x\in D \cap B} f(x) \leq \sum_{x\in D} f(x)=1$since $f$ is non-negative. Clearly the minimum value of $\mathbb{P}(B)$ is zero which it attains when $D \cap B = \emptyset.$ Therefore $\mathbb{P}(B)\in [0,1].$ (2) By the assumption that $f$ is a probability mass function, we have $\mathbb{P}(\mathbb{R})=\sum_{x\in D \cap \mathbb{R}} f(x)=\sum_{x\in D}f(x)=1.$(3) Let $(A_{n})_{n\geq 1}$ such that $A_{n}\in \mathcal{B}(\mathbb{R})$ for all $n\geq 1$ and $A_{n}\cap A_{m}=\emptyset$ for all $n,m \geq 1$ such that $n \neq m.$ Then $\mathbb{P}\left( \bigcup_{i\geq 1} A_{i} \right) = \sum_{x\in \left( \bigcup_{i \geq 1}A_{i} \right) \cap D} f(x) = \sum_{x\in \bigcup_{i \geq 1}(A_{i} \cap D)} f(x)= \sum_{i \geq 1} \sum_{x\in A_{i}\cap D} f(x) = \sum_{i \geq 1} \mathbb{P}(A_{i}). $ Now $X$ is a [[Random Variables|real-valued random variable]] since for all $a\in \mathbb{R},$ $X^{-1}((-\infty,a])=(-\infty ,a]\in \mathcal{F}=\mathcal{B}(\mathbb{R}).$ And $X$ is discrete since $D$ is countable and $\mathbb{P}(\mathbb{R}\setminus D)=\sum_{x \in D \cap (\mathbb{R}\setminus D)} f(x)=0.$ Lastly, indeed $p_{X}=f$: let $x\in \mathbb{R}$ then $p_{X}(x)=\mathbb{P}_{X}(\{ x \})=\mathbb{P}(X)= \sum_{z\in \{ x \} \cap D}f(z)= \sum_{z\in\{ x \}}f(z)=f(z)$