Theorem 16.13. If f is nonnegative, then

$${\int}_{\mathrm{\Omega}}f(T\omega )\mu (d\omega )={\int}_{{\mathrm{\Omega}}^{\mathrm{\prime}}}f\left({\omega}^{\mathrm{\prime}}\right)\mu {T}^{-1}\left(d{\omega}^{\mathrm{\prime}}\right).$$

A function f (not necessarily nonnegative) is integrable with respect to $\mu {T}^{-1}$ if and only if fT is integrable with respect to $\mu $, in which case (16.17) and

$${\int}_{{T}^{-1}{A}^{\mathrm{\prime}}}f(T\omega )\mu (d\omega )={\int}_{{A}^{\mathrm{\prime}}}f\left({\omega}^{\mathrm{\prime}}\right)\mu {T}^{-1}\left(d{\omega}^{\mathrm{\prime}}\right)$$

hold. For nonnegative f, (16.18) always holds.

(2.47) Theorem. Suppose $\mathrm{\Omega}$ is an open set in ${\mathbf{R}}^{n}$ and $G:\mathrm{\Omega}\to {\mathbf{R}}^{n}$ is a ${C}^{1}$ diffeomorphism. (a) If f is a Lebesgue measurable function on $G(\mathrm{\Omega})$, then $f\circ G$ is Lebesgue measurable on $\mathrm{\Omega}$. If $f\ge 0$ or $f\in {L}^{1}(G(\mathrm{\Omega}),m)$, then

$${\int}_{G(\mathrm{\Omega})}f(x)dx={\int}_{\mathrm{\Omega}}f\circ G(x)|\mathrm{det}{D}_{x}G|dx$$

(b) If $E\subset \mathrm{\Omega}$ and $E\in {\mathcal{L}}^{n}$, then $G(E)\in {\mathcal{L}}^{n}$ and $m(G(E))=$ ${\int}_{E}|\mathrm{det}{D}_{x}G|dx$.

$${\int}_{\mathrm{\Omega}}f(T\omega )\mu (d\omega )={\int}_{{\mathrm{\Omega}}^{\mathrm{\prime}}}f\left({\omega}^{\mathrm{\prime}}\right)\mu {T}^{-1}\left(d{\omega}^{\mathrm{\prime}}\right).$$

A function f (not necessarily nonnegative) is integrable with respect to $\mu {T}^{-1}$ if and only if fT is integrable with respect to $\mu $, in which case (16.17) and

$${\int}_{{T}^{-1}{A}^{\mathrm{\prime}}}f(T\omega )\mu (d\omega )={\int}_{{A}^{\mathrm{\prime}}}f\left({\omega}^{\mathrm{\prime}}\right)\mu {T}^{-1}\left(d{\omega}^{\mathrm{\prime}}\right)$$

hold. For nonnegative f, (16.18) always holds.

(2.47) Theorem. Suppose $\mathrm{\Omega}$ is an open set in ${\mathbf{R}}^{n}$ and $G:\mathrm{\Omega}\to {\mathbf{R}}^{n}$ is a ${C}^{1}$ diffeomorphism. (a) If f is a Lebesgue measurable function on $G(\mathrm{\Omega})$, then $f\circ G$ is Lebesgue measurable on $\mathrm{\Omega}$. If $f\ge 0$ or $f\in {L}^{1}(G(\mathrm{\Omega}),m)$, then

$${\int}_{G(\mathrm{\Omega})}f(x)dx={\int}_{\mathrm{\Omega}}f\circ G(x)|\mathrm{det}{D}_{x}G|dx$$

(b) If $E\subset \mathrm{\Omega}$ and $E\in {\mathcal{L}}^{n}$, then $G(E)\in {\mathcal{L}}^{n}$ and $m(G(E))=$ ${\int}_{E}|\mathrm{det}{D}_{x}G|dx$.