Probability Distribution is Probability Measure

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Theorem

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.

Let $\struct {S, \Sigma'}$ be a measurable space.

Let $X$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$.

Let $P_X$ be the probability distribution of $X$.


Then:

$P_X$ is a probability measure on $\struct {S, \Sigma'}$.


Proof

From the definition of probability distribution, we have:

$P_X = X_* \Pr$

where $X_* \Pr$ is the pushforward $X_* \Pr$ of $\Pr$, under $X$, on $\Sigma'$.

From Pushforward Measure is Measure, we have:

$P_X$ is a measure.

We then have:

\(\ds \map {P_X} S\) \(=\) \(\ds \map \Pr {X^{-1} \sqbrk S}\) Definition of Probability Distribution
\(\ds \) \(=\) \(\ds \map \Pr \Omega\)
\(\ds \) \(=\) \(\ds 1\) Definition of Probability Measure

so:

$P_X$ is a probability measure.

$\blacksquare$