Integral with respect to Pushforward Measure

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $\left({X', \Sigma'}\right)$ be a measurable space.

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

Let $f: X' \to \overline \R$ be a $T \left({\mu}\right)$-integrable function, where $T \left({\mu}\right)$ denotes the pushforward measure of $\mu$ under $T$.


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

$\displaystyle \int_{X'} f \, \mathrm d T \left({\mu}\right) = \int_X f \circ T \, \mathrm d \mu$


Proof


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