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:


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

Proof
Let $S \in \Sigma'$ be arbitrary.

We have:

By linearity, we obtain:
 * $\ds (1): \int_{X'} f \rd \map T \mu = \int_X f \circ T \rd \mu$

for simple $f: X' \to \hointr 0 \infty$.

Now let $f: X' \to \closedint 0 \infty$ be an arbitrary measurable map.

There is an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of simple functions $f_n: X' \to \hointr 0 \infty$ that converges pointwise to $f$.

Note that $\sequence {f_n \circ T}_{n \mathop \in \N}$ is also an increasing sequence and converges pointwise to $f \circ T$.

Then:

Now suppose $f: X' \to \overline \R$ is $\map T \mu$-integrable.

Then $f^+, f^-: X' \to \closedint 0 \infty$ are both $\map T \mu$-integrable.

Thus by the previous result: