> [!definition] > > Let $X$ be a [[Random Variable|random variable]] on a [[Probability|probability space]] $(\Omega, \cf, P)$, and $\cb$ be a [[Sigma Algebra|sigma algebra]] on $\real$ (typically taken to be the [[Borel Sigma Algebra|Borel sigma algebra]]). If $X$ is $(\cf, \cb)$-[[Measurable Function|measurable]], then > $ > \mathcal{L}_X = \mu_X: \cb \to [0, 1] \quad B \mapsto P(X^{-1}(B)) > $ > is a probability [[Measure Space|measure]] on $(\real, \cb)$, known as the **probability distribution** of $X$. > > *Proof*. The preimage preserves unions, intersections, and complements[^2]. Any disjoint sequence of sets $\seq{B_n}$ also satisfies $\seq{X^{-1}(B_n)}$ disjoint. So $\mu_X(\emptyset) = P(\emptyset) = 0$, and > $ > \begin{align*} > \mu_X\paren{\bigcup_{n \in \nat}B_n} > &= P\paren{X^{-1}\paren{\bigcup_{n \in \nat}B_n}}\\ > &= P\paren{\bigcup_{n \in \nat}X^{-1}(B_n)} \\ > &= \sum_{n \in \nat}P(X^{-1}(B_n)) \\ > &= \sum_{n \in \nat}\mu_X(B_n) > \end{align*} > $ > we have $\mu_X$ being a measure. Lastly, $\mu_X(\real) = P(\Omega) = 1$. Therefore $\mu_X$ is a probability measure. [^1]: The $F(b) - F(a)$ needs to be adjusted to also include $a$. The [[Summation|sum]] covers this. [^2]: See [[Measurable Function|measurable function]].