Integrable Function under Pushforward Measure

From ProofWiki
Jump to navigation Jump to search

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 mapping.


Then the following are equivalent:

$(1): \quad f$ is $T \left({\mu}\right)$-integrable
$(2): \quad f \circ T$ is $\mu$-integrable

where $T \left({\mu}\right)$ is the pushforward measure of $\mu$ under $T$.


Proof


Sources