Equivalent way of writing the norm of Lp Given a measurable $E\subset \Bbb R^d $  and a measurable function $f:E\rightarrow \Bbb R^d $, prove that :
$$
\int (\left\lvert f \right\rvert)^r d\mu  = r\int_{0}^\infty t^{r-1} \mu(\{x  \in  E \mid  \left\lvert f(x) \right\rvert>t\})\,dt 
$$
where $r   \ge  1$
The $r=1$ case is simple; however, I can't prove the general case. Thank you in advance
 A: If you know how to prove the case $r=1$, it's easy :
let $g(t)= |f(t)|^r$. You have (assuming you can prove the result for $r=1$)
$$\int_E |f(t)|^r dt = \int_E |g(t)| dt = \int_0^{\infty} \mu(\{ x\in E \ :\ |g(x)| > t\}) dt$$
it follow
$$= \int_0^{\infty} \mu(\{ x\in E \ :\ |f(x)|^r > t\}) dt = \int_0^{\infty} \mu(\{ x\in E \ :\ |f(x)| > t^{\frac{1}{r}}\}) dt $$
Now, you do the change of variable $y= t^\frac{1}{r}$, and it gives you the result :
$$= \int_0^{\infty} ry^{1-r} \mu(\{ x\in E \ :\ |f(x)| > y \}) dy $$
A: The basic idea is to use $t^r = r \int_0^t s^{r-1} ds$ combined with Fubini Tonelli.
Suppose $g$ is a non negative measurable function.
Then
\begin{eqnarray}
\int g(x)^r d \mu(x) &=& \int r \int_0^{g(x)}s^{r-1}ds d \mu(x) \\
&=& \int r \int_0^\infty 1_{[0,g(x))}(s)s^{r-1}ds d \mu(x) \\
&=& r \int_0^\infty s^{r-1} \left( \int 1_{[0,g(x))}(s) d \mu(x) \right) ds \\
&=& r \int_0^\infty s^{r-1} \left( \int 1_{\{y | g(y) > s \}}(x) d \mu(x) \right) ds \\
&=& r \int_0^\infty s^{r-1} \mu( \{y | g(y) > s \} ) ds
\end{eqnarray}
A: You can derive the general case from the case $r=1$. Because $|f|>t$ if and only if $|f|^r>t^r$.
