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Let $\mu$ be a (positive) measure on $\mathbb{R}^d$ and $f$ be a $\mu$-measurable function on $\mathbb{R}^d$. How to prove that \begin{equation} \int_{\mathbb{R}^d} |f(x)|^p d\mu(x)=p\int_{0}^{\infty} \gamma^{p-1} \mu(\{x\in \mathbb{R}^d:|f(x)|>\gamma\}) d\gamma \end{equation} for every $1\leq p<\infty$ and every $\gamma>0$.

I found in Folland book that we have to prove the following equation:

\begin{align} \int_{\mathbb{R}^d} \phi(|f(x)|) d\mu(x)= -\int_{0}^{\infty} \phi(\gamma) d(\mu(\{x\in \mathbb{R}^d: |f(x)|>\gamma\})) \end{align} for non-negative function $\phi$ and using integration by part to get the result. Could we get some direct proof?

beginner
  • 2,225

1 Answers1

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You can do it using the Indicator-Fubini trick. Here's how it goes:

$$\int_{\mathbb{R}^d}|f(x)|^pd\mu(x) = \int_{\mathbb{R}^d}\int_0^{|f(x)|}pu^{p-1} dud\mu(x)$$ $$ = \int_{\mathbb{R}^d}\int_0^{\infty}1_{\{u < |f(x)|\}}pu^{p-1} dud\mu(x)$$

By Fubini Theorem,

$$ = \int_0^{\infty}pu^{p-1}\int_{\mathbb{R}^d}1_{\{u < |f(x)|\}} d\mu(x)du$$ $$ = \int_0^{\infty}pu^{p-1}\mu\{x \in \mathbb{R^d}: |f(x)| > u \}du$$

$\blacksquare$

Gautam Shenoy
  • 10,598