Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces, let $\mu$ be a probability measure on $(X,\mathcal{A})$, let $\nu$ be a probability measure on $(Y,\mathcal{B})$, and let $T: X \to Y$ be an $\mathcal{A}$-$\mathcal{B}$ measurable map. Then $T_{\#}\mu = \nu$ if and only if, for every bounded $\mathcal{B}$-measurable function $f: Y \to \mathbb{R}$, the composition $f \circ T: X \to \mathbb{R}$ is $\mathcal{A}$-measurable and

\begin{align*} \int_Y f(y)\, d\nu(y) = \int_X f(T(x))\, d\mu(x). \end{align*}