Pushforward Measure is Measure

Theorem
Let $\left({X, \Sigma}\right)$ and $\left({X', \Sigma'}\right)$ be measurable spaces.

Let $\mu$ be a measure on $\left({X, \Sigma}\right)$.

Let $f: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.

Then the pushforward measure $f_* \mu: \Sigma' \to \overline{\R}$ is a measure.