How to show $\int_0^\infty Vol( \{x\in M : \phi(x)\ge t \} \cap M) dt =\int_M \phi $?

Question

$M$ is a open set of $R^n$ with regular boundary . $\phi :M \rightarrow R^+$ is smooth function . How to show $$ \int_0^\infty Vol( \{x\in M : \phi(x)\ge t \} \cap M) dt =\int_M \phi $$

Answer 1

It's a simple consequence of Tonelli theorem. I denote $Vol$ by $\mu$ $$\begin{align} \int_0^\infty \mu( \{x\in M : \phi(x)\ge t \} \cap M) dt &=\int_0^\infty \mu( \{x\in M : \phi(x)\ge t \} ) dt \\ &= \int_0^\infty \int \chi_{\{x\in M : \phi(x)\ge t \}}(y) d\mu(y) dt \\ &= \int \int_0^\infty \chi_{\{x\in M : \phi(x)\ge t \}}(y) dt d\mu(y) \\ &= \int \int_0^\infty \chi_{M}(y) \chi_{[0,\phi(y)]}(t) dt d\mu(y)\\ &= \int \chi_{M}(y) \int_0^\infty\chi_{[0,\phi(y)]}(t) dt d\mu(y)\\ &= \int \chi_{M}(y) \phi(y) d\mu(y)\\ &= \int_M \phi(y) d\mu(y) \end{align}$$