${\displaystyle {}(M,{\mathcal {A}})}$

${\displaystyle {}{\mathcal {E}}}$

${\displaystyle {}{\mathcal {A}}}$

${\displaystyle {}\mu _{1}}$

${\displaystyle {}\mu _{2}}$

${\displaystyle {}(M,{\mathcal {A}})}$

${\displaystyle {}{\mathcal {E}}}$

${\displaystyle {}M_{n}\uparrow M}$

${\displaystyle {}M_{n}\in {\mathcal {E}}}$

${\displaystyle {}\mu _{1}(M_{n})=\mu _{2}(M_{n})<\infty }$

${\displaystyle {}\mu _{1}=\mu _{2}\,.}$

${\displaystyle {}T\subseteq M\times N}$

${\displaystyle {}(\mu \otimes \nu )(T)=\int _{M}\nu (T(x))\,d\mu (x)=\int _{N}\mu (T(y))\,d\nu (y)\,.}$

${\displaystyle f\colon H\longrightarrow \mathbb {R} }$

ist ${\displaystyle {}f}$ genau dann integrierbar auf ${\displaystyle {}H}$, wenn die Hintereinanderschaltung ${\displaystyle {}f\circ \varphi }$ auf ${\displaystyle {}G}$ integrierbar ist. In diesem Fall gilt

${\displaystyle {}\int _{H}f\,d\lambda ^{n}=\int _{G}(f\circ \varphi )\vert {J(\varphi )}\vert \,d\lambda ^{n}\,,}$

${\displaystyle {}J(\varphi )}$

${\displaystyle {}D\varphi }$

${\displaystyle {}M}$

${\displaystyle {}n}$

${\displaystyle {}\partial M}$

${\displaystyle {}\omega }$

${\displaystyle {}(n-1)}$

${\displaystyle {}M}$

${\displaystyle {}\int _{M}d\omega =\int _{\partial M}\omega \,.}$