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 positive $\Sigma'$-measurable function.

Let $\map T \mu$ be the pushforward measure of $\mu$ under $T$.

Then $f \circ T: X \to \overline \R$ is positive and $\Sigma$-measurable with:


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

Proof
From Composition of Measurable Mappings is Measurable:


 * $f \circ T$ is $\Sigma$-measurable.

Clearly $f \circ T \ge 0$, so $f \circ T$ is a positive $\Sigma$-measurable function.

We first show the proposition for characteristic functions $f$.

Suppose $f = \chi_A$ for $A \in \Sigma'$.

Then, we have:

So, we have:


 * $\ds \int_{X'} \chi_A \rd \map T \mu = \int_X \chi_A \circ T \rd \mu$

whenever $A \in \Sigma'$.

Now suppose that $f$ is a positive simple function.

From Simple Function has Standard Representation, there exists:


 * a finite sequence $a_0, \ldots, a_n$ of non-negative real numbers
 * a partition $E_0, E_1, \ldots, E_n$ of $X$ into $\Sigma'$-measurable sets

such that:


 * $\ds f = \sum_{i \mathop = 0}^n a_i \chi_{E_i}$

Then:

Now suppose that $f$ is a general positive $\Sigma'$-measurable function.

From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that:


 * $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

for each $x \in X$.

For each $n \in \N$, we have:


 * $\ds \int_{X'} f_n \rd \map T \mu = \int_X f_n \circ T \rd \mu$

Then: