Integral with respect to Pushforward Measure

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

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

Let $T: X \to X'$ be a $\Sigma \, / \, \Sigma'$-measurable mapping.

Let $f: X' \to \overline \R$ be a $\map T \mu$-integrable function, where $\map T \mu$ denotes the pushforward measure of $\mu$ under $T$.

Then $f \circ T: X \to \overline \R$ is $\mu$-integrable, and:


 * $\displaystyle \int_{X'} f \rd \map T \mu = \int_X f \circ T \rd \mu$

Proof
Let $S \in \Sigma'$ be arbitrary. We have \begin{align} \int_{X'} \chi_S \,dT(\mu) &= T(\mu)(S) \\ &= \mu(T^{-1}(S)) \\ &= \int_{X} \chi_{T^{-1}(S)} \,d\mu \\ &= \int_{X} \chi_S \circ T \,d\mu. \end{align} By linearity, we obtain $\int_{X'} f \,dT(\mu) = \int_{X} f \circ T \,d\mu$ for simple $f \colon X' \to [0, \infty)$. Now let $f \colon X' \to [0, \infty]$ be an arbitrary measurable map. There is a sequence $(f_n)_{n = 1}^{\infty}$ of simple functions $f_n \colon X' \to [0, \infty)$ such that $f_n(x) \nearrow f(x)$ for each $x \in X'$. Note that $f_n(T(x)) \nearrow f(T(x))$ for each $x \in X$. By the monotone convergence theorem and the result established for nonnegative simple functions, \begin{align} \int_{X'} f \,dT(\mu) &= \lim_{n \to \infty}\int_{X'} f_n \,dT(\mu) \\ &= \lim_{n \to \infty} \int_{X} f_n \circ T \,d\mu \\ &= \int_{X} f \circ T \,d\mu. \end{align}

Now suppose $f \colon X' \to \overline{\mathbb{R}}$ is $T(\mu)$-integrable. Then $f^+, f^- \colon X' \to [0, \infty]$ are both $T(\mu)$-integrable. Thus by the previous result, \begin{align} \int_{X'} f\,dT(\mu) &= \int_{X'}f^+ \,dT(\mu) - \int_{X'}f^- \,dT(\mu) \\ &= \int_{X} f^+ \circ T \,d\mu - \int_{X} f^- \circ T \,d\mu \\ &= \int_{X} (f^+ \circ T - f^- \circ T) \,d\mu \\ &= \int_{X} f \circ T \,d\mu. \end{align}