Relation between integral and distribution 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?

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