A measurable function from a first measure space to a second measurable space induces a measure on the latter.
Consider two measurable spaces and a measurable function between them. The image measure of a measure on the first space under the measurable function is the measure on the second space which assigns to each measurable set (of that space) the measure of the inverse image of that set.
We say that the function induces the image measure on the codomain. Alternatively, we say that we push forward the measure to the codomain, and so call the image measure a push forward measure.
Let
$(X, \mathcal{A} )$
and
$(Y, \mathcal{B} )$
be two measurable spaces.
Let $f: X \to Y$ be
a measurable function.
Let
$\mu : \mathcal{A} \to \nneri$
be a measure.
We denote the
image measure of $\mu $
under $f$ by
$\pushforward{\mu }{f}$,
for the reason that it
\[
\pushforward{\mu }{f}(B) = \mu (f^{-1}(B))
\]
The main property we would like to hold is that integration on the new measure space is the same as integration on the old.
Then $g: Y \to $ is integrable with respect
to $\pushforward{\mu }{f}$ if and only if $g
\circ f$ is integrable with respect to $\mu $.
In this case,
\[
\int g d(\pushforward{\mu }{f})
=
\int g \circ f d\mu .
\]