$\int_X F\operatorname{d} \mu=\int_Y F\circ g\operatorname{d} \nu $ where $g$ is invertible and $ \nu(B)=\mu(g B)$

Suppose $g$ is a measurable map from a measure space $(X,\mathscr F,\mu)$ to a measurable space $(Y,\mathscr S)$. Define $$\nu(B)=\mu(g^{-1} B) \forall B\in \mathscr S,$$ then $(Y,\mathscr S,\nu)$ is a measure space, and for any measurable function $f$ on $Y$, the following equation holds as long as one side makes sense:$\newcommand{\d}{\operatorname{d}}$
$$\int_Y f\d \nu = \int_X f\circ g\d \mu. $$

In particular, if $g$ is invertible and $h=g^{-1}$, then $$\nu(B)=\mu(h B)\forall B\in \mathscr S,$$ and for any measurable function $F$ on $X$, the following equation holds as long as one side makes sense:$\newcommand{\d}{\operatorname{d}}$
$$\int_X F\d \mu=\int_Y F\circ h\d \nu . $$

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