Prove that $\int_E |g|^2 \mathop{d \mu} = 2 \int_0^\infty x \, \mu(|g| > x) \mathop{dx}$

Question

Let $(E,\mu)$ be a measure space.

Why for all $g \in L^2(\mu)$, we have $$\int_E |g|^2 \mathop{d \mu} = 2 \int_0^\infty x \, \mu(|g| > x) \mathop{dx} ?$$

Answer 1

It is known as Robin's identity in probability theory. In fact, we can generalize it to the following:

Let $p\in[1,\infty)$ and let $(E,\mathcal{M},\mu)$ be a measure space. Let $f:E\rightarrow[0,\infty)$ be a measurable function such that $\int f^{p}\,d\mu<\infty$. Then \begin{eqnarray*} \int f^{p}\,d\mu & = & \int_{0}^{\infty}pt^{p-1}\mu\left(f^{-1}\left[[t,\infty)\right]\right)dt.\\ & = & \int_{0}^{\infty}pt^{p-1}\mu\left(f^{-1}\left[(t,\infty)\right]\right)dt \end{eqnarray*}

Proof: Since $\int f^{p}\,d\mu<\infty$, by restricting $\mu$ on the support of $f$, without loss of generality, we may assume that $\mu$ is $\sigma$-finite. Let $m$ be the Lebesgue measure on $\mathbb{R}$. Define a set $A=\{(t,\omega)\mid0<t<f(\omega)\}\subseteq\mathbb{R}\times E$. Firstly, we show that $A$ is measurable with respect to the product $\sigma$-algebra. We assert that $$ A=\bigcup_{\substack{a,b\in\mathbb{Q}\\ 0<a<b } }(a,b)\times f^{-1}\left[(b,\infty)\right]. $$ For, let $(t,\omega)\in A$, then $0<t<f(\omega)$. There exist $a_{0},b_{0}\in\mathbb{Q}$ such that $0<a_{0}<t<b_{0}<f(\omega)$. Therefore $$ (t,\omega)\in(a_{0},b_{0})\times f^{-1}\left[(b_{0},\infty)\right]\subseteq\bigcup_{\substack{a,b\in\mathbb{Q}\\ 0<a<b } }(a,b)\times f^{-1}\left[(b,\infty)\right]. $$ Conversely, if $(t,\omega)\in\bigcup_{\substack{a,b\in\mathbb{Q}\\ 0<a<b } }(a,b)\times f^{-1}\left[(b,\infty)\right]$, then there exist $a_{0},b_{0}\in\mathbb{Q}$ with $0<a_{0}<b_{0}$ such that $(t,\omega)\in(a_{0},b_{0})\times f^{-1}\left[(b_{0},\infty)\right]$. Therefore $0<a_{0}<t<b_{0}<f(\omega)$ and hence $(t,\omega)\in A$. It follows that $$ A=\bigcup_{\substack{a,b\in\mathbb{Q}\\ 0<a<b } }(a,b)\times f^{-1}\left[(b,\infty)\right]. $$ Note that for each $a,b\in\mathbb{Q}$ with $0<a<b$, $(a,b)\times f^{-1}\left[(b,\infty)\right]$ is a measurable subset (with respect to the product $\sigma$-algebra) of $\mathbb{R}\times E$. Since the union is countable, $A$ is measurable with respect to the product $\sigma$-algebra.

Now, we are ready to apply Fubini-Tonelli Theorem. Consider the integral with respect to the product measure $m\times\mu$ $$ \int ps^{p-1}1_{A}(s,\omega)\,d(m\times\mu)(s,\omega). $$ On one hand, \begin{eqnarray*} & & \int ps^{p-1}1_{A}(s,\omega)\,d(m\times\mu)(s,\omega)\\ & = & \int\left(\int ps^{p-1}1_{A}(s,\omega)dm(s)\right)d\mu(\omega)\\ & = & \int\left(\int_{0}^{f(\omega)}ps^{p-1}dm(s)\right)d\mu(\omega)\\ & = & \int\left(f(\omega)\right)^{p}\,d\mu(\omega). \end{eqnarray*} On the other hand, \begin{eqnarray*} & & \int ps^{p-1}1_{A}(s,\omega)\,d(m\times\mu)(s,\omega)\\ & = & \int\left(\int ps^{p-1}1_{A}(s,\omega)d\mu(\omega)\right)dm(s)\\ & = & \int_{0}^{\infty}\left(ps^{p-1}\mu\left(f^{-1}\left[(s,\infty)\right]\right)\right)dm(s). \end{eqnarray*} Therefore $$ \int f^{p}d\mu=p\int_{0}^{\infty}s^{p-1}\mu\left(f^{-1}\left[(s,\infty)\right]\right)ds. $$ Lastly, by replacing $A$ with $A=\{(t,\omega)\mid0\leq t\leq f(\omega)\}$ and repeat the same argument, we can show that $$ \int f^{p}d\mu=p\int_{0}^{\infty}s^{p-1}\mu\left(f^{-1}\left[[s,\infty)\right]\right)ds. $$