Why is $\int_{-\infty}^\infty |g(x)|\,dx = \int_0^\infty \mu(\{x : |g(x)| \ge t\})\,dt$ true?

Question

Why do we have the following equality$$\int_{-\infty}^\infty |g(x)|\,dx = \int_0^\infty \mu(\{x : |g(x)| \ge t\})\,dt,$$where $\mu$ is Lebesgue measure?

Answer 1

We claim that for a nonnegative measurable function $g:\Bbb{R}\to[0,\infty)$, $$ \int_\Bbb{R}g(x)\ d\mu(x)=\int_{[0,\infty)}\mu(\{x\in\Bbb{R}\mid g(x)\geq s\}) \ d\mu(s). $$ This is a good example of applications of the Fubini-Tonelli's Theorem.

Let $\nu:=g_*\mu$ be the pushforward of $\mu$, i.e., $\nu=\mu\circ g^{-1}$. Then $$ \int_\Bbb{R}g(x)\ d\mu(x)=\int_{[0,\infty)}x\ d\nu(x). $$ Note that $$ \begin{align*} \int_{[0,\infty)}x\ d\nu(x)&=\int_{[0,\infty)}\left(\int_{[0,\infty)}1_{[0,x]}(y)\ d\mu(y)\right)\ d\nu(x)\\ &=\int_{[0,\infty)} \left(\int_{[0,\infty)}1_{[y,\infty]}(x)\ d\nu(x)\right)\ d\mu(y)\\ &=\int_{[0,\infty)} \nu([y,\infty))\ d\mu(y)\\ &=\int_{[0,\infty)} \mu\circ g^{-1}([y,\infty))\ d\mu(y)\\ &=\int_{[0,\infty)}\mu(\{x\in\Bbb{R}\mid g(x)\geq y\}) \ d\mu(y). \end{align*} $$

Answer 2

Observe \begin{align} \int^\infty_0 \mu\{x \mid |g(x)|\geq t\}\ dt =& \int^\infty_0 \int^\infty_{-\infty} \chi_{\{x \in \mathbb{R} \mid |g(x)| \geq t\}}\ dxdt\\ =&\ \int^\infty_{-\infty} \int^\infty_0\chi_{\{x \in \mathbb{R} \mid |g(x)| \geq t\}}\ dtdx\\ =&\ \int^\infty_{-\infty} \int^{|g(x)|}_0 \ dtdx \\ =&\ \int^\infty_{-\infty} |g(x)|\ dx. \end{align}

Edit: Note that the interchanging of integrals are allowed because everything is non-negative.