Spherical rearrangement Let $u\colon\Omega\subset\mathbb{R}^N\to\mathbb{R}$ be a non negative measurable function, and $\Omega$ open and bounded.
 Consider $u^*$ the spherical rearrangement 
$$
u^*(x)=\sup\{t\geq0 : \mu\{x: u(x)\geq t \} > \omega_N |x|^N\}
$$
where $\mu$ is the Lebesgue measure in $\mathbb{R}^N$, $w_N=\mu\{B(0,1)\}$ of the unitary ball in $\mathbb{R}^N$ (i.e. to the definition in http://www.math.toronto.edu/almut/rearrange.pdf). 
Consider a continuous function $F\colon \mathbb{R}\to \mathbb{R}$, then prove
\begin{equation}\tag{1}
\int_\Omega{F(u(x))dx}=\int_\Omega{F(u^*(x))dx}.
\end{equation}
I try to consider 
$$
\int_\Omega{F(u(x))dx}=\int_{-\infty}^{+\infty}{\mu\{x: F(u(x))\geq t \}dt}=\int_{-\infty}^{+\infty}{\mu\{x: F(u(x))\geq F(s) \}F'(s)ds}
$$
but in this case i need a hyphotesis of derivability. Also (1) is said to be true with $\Omega=\mathbb{R}^N$, but i think that is necessary that $F\in L^1(\mathbb{R}^N)$.
 A: I think you're on the right track, but you need the following ingredient:

For almost all $a,  b\in \mathbb{R}$ with $a<b$,
   $$
 \mu\{x; u(x)\in (a,b)\} = \mu\{x; u^*(x) \in (a,b)\}.
 $$

To prove this, observe that, for almost all $a, b$, $u^*(x) \in (a,b)$ is equivalent to:
$$
\mu\{x; u(x)\geq b\} \leq \omega_N |x|^N \leq \mu\{x; u(x)\geq a\}.
$$
This defines an annulus of measure $\mu\{x; u(x)\geq a\} - \mu\{x; u(x)\geq b\}$, which is for almost all $a,b$ equal to the left hand side of our claim.  Specifically, so long as $\{x; u(x) = a\}$ and $\{x; u(x) = b\}$ are sets of measure zero, we're fine.
It's not hard to get from here to 
$$
\mu\{x; F(u(x)) > t\} = \mu\{x; F(u^*(x)) > t\}
$$
for almost all $t$.  Indeed since $F$ is continuous, $F^{-1}((t,\infty))$ is open and thus a union of open intervals.  I think you can work with weaker hypotheses, but I'm not completely sure.  For example, if $F$ had a discontinuity at some $a$ where $\mu\{x; u(x) = a\} > 0$, perhaps there could be a problem?  My intuition says that $F$ being measurable should be fine though.
Anyway, you can complete the proof just how you started it:
\begin{align*}
\int_{\Omega} F(u(x))\,dx &= 
\int_{-\infty}^\infty \mu\{x; F(u(x)) > t\} \,dt \\
&= \int_{-\infty}^\infty \mu\{x; F(u^*(x)) > t\} \,dt \\
&= \int_{\Omega} F(u^*(x))\,dx.
\end{align*}
Like I mentioned in the comment, I don't think this works when $\Omega$ is not a ball centered at the origin.  Also, if $\Omega = \mathbb{R}^N$, then I believe what you need is $F\circ u\in L^1$, which depends on the behavior of $F(t)$ as $t\to\infty$ and $t\to 0$, as well as the conditions on $u$.
