Let $(X,\mathcal A,\mu)$ be a [measure space](/page/Measure%20Space), let $(Y,\mathcal B)$ be a measurable space, and let $T:X\to Y$ be an $\mathcal A/\mathcal B$-measurable map. Let $T_{\#}\mu$ denote the pushforward measure on $(Y,\mathcal B)$, defined by $(T_{\#}\mu)(B)=\mu(T^{-1}(B))$ for every $B\in\mathcal B$.

If $f:Y\to[0,\infty]$ is $\mathcal B$-measurable, then $f\circ T:X\to[0,\infty]$ is $\mathcal A$-measurable and

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

If $f:Y\to\mathbb R$ is $\mathcal B$-measurable and the signed integral of either $f$ with respect to $T_{\#}\mu$ or $f\circ T$ with respect to $\mu$ is well-defined, then both signed integrals are well-defined and the same identity holds. In particular, this applies whenever either $f\in L^1(Y,\mathcal B,T_{\#}\mu)$ or $f\circ T\in L^1(X,\mathcal A,\mu)$ holds.

If $f:Y\to\mathbb C$ is $\mathcal B$-measurable and the signed integrals of the real and imaginary parts are well-defined on either side, then the corresponding component integrals are well-defined on both sides and the same identity holds. In particular, this applies whenever either $f\in L^1(Y,\mathcal B,T_{\#}\mu)$ or $f\circ T\in L^1(X,\mathcal A,\mu)$ holds.