Integral with respect to Pushforward Measure/Corollary

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

In this case, we have:


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

Proof
From Function Measurable iff Positive and Negative Parts Measurable, we have:


 * $f^+$ and $f^-$ are $\Sigma'$-measurable.

From Composition of Measurable Mappings is Measurable, we have:


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

Then from Function Measurable iff Positive and Negative Parts Measurable, we have:


 * $\paren {f \circ T}^+$ and $\paren {f \circ T}^-$ are $\Sigma$-measurable.

Then, we have:

and:

So:


 * $\ds \int_X \paren {f \circ T}^+ \rd \mu < \infty$ and $\ds \int_X \paren {f \circ T}^- \rd \mu <\infty$




 * $\ds \int_{X'} f^+ \rd \map T \mu < \infty$ and $\ds \int_{X'} f^- \rd \map T \mu < \infty$.

So $f \circ T : X \to \overline \R$ is $\mu$-integrable $f : X' \to \overline \R$ is $\map T \mu$-integrable

In this case: