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 :

$$
\int (\left\lvert f \right\rvert)^r d\mu  = r\int_{0}^\infty t^{r-1} \mu(\{x  \in  E \mid  \left\lvert f(x) \right\rvert&gt;t\})\,dt 
$$
where $r   \ge  1$

The $r=1$ case is simple; however, I can't prove the general case. Thank you in advance
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 $r=1$)
$$\int_E |f(t)|^r dt = \int_E |g(t)| dt = \int_0^{\infty} \mu(\{ x\in E \ :\ |g(x)| &gt; t\}) dt$$

it follow

$$= \int_0^{\infty} \mu(\{ x\in E \ :\ |f(x)|^r &gt; t\}) dt = \int_0^{\infty} \mu(\{ x\in E \ :\ |f(x)| &gt; t^{\frac{1}{r}}\}) dt $$

Now, you do the change of variable $y= t^\frac{1}{r}$, and it gives you the result :

$$= \int_0^{\infty} ry^{1-r} \mu(\{ x\in E \ :\ |f(x)| &gt; y \}) dy $$
You can derive the general case from the case $r=1$. Because $|f|&gt;t$ if and only if $|f|^r&gt;t^r$.