Schwarz symmetrization is equimeasurable Suppose $\Omega\subset\mathbb{R^2}$ is open and bounded, and let $f:\Omega\rightarrow [0,\infty)$ be measurable. Moreover, let $\Omega^{\ast}$ denote the closed disk with midpoint $0\in\mathbb{R}^2$ and the same area as $\Omega$ (i.e. $\vert\Omega\vert=\vert \Omega^{\ast} \vert$). 
The Schwarz symmetrization is defined as the function 
$$f^{\ast}:\Omega^{\ast}\rightarrow[0,\infty),\quad f(x):=\sup\lbrace{ c\geq 0:x\in \Omega_c^{\ast} \rbrace},$$
where $\Omega_c:=f^{-1}([c,\infty))\subset\Omega$ and $\Omega_c^{\ast} $ denotes the closed disk with midpoint $0\in\mathbb{R}^2$ and $\vert\Omega_c^{\ast} \vert= \vert\Omega_c \vert$.
I would like to understand the following two problems:
1) Why are $f$ and $f^{\ast}$ equimeasurable, i.e. $\vert \lbrace x\in\Omega^{\ast}: f^{\ast}(x)\geq c\rbrace \vert=\vert \Omega_{c} \vert$ for all $c\geq 0$ ? 
2) Sometimes, the function $f^{\ast}$ is defined alternatively as 
$$f^{\ast}(x):=\int_{0}^{\infty} \chi_{ {\lbrace f \geq t \rbrace}^{\ast} }(x) dt.$$
I am wondering if that definition is equivalent to the first one?
Unfortunately, I do not know how to answer 1) or 2). I think that if we use $\vert\Omega_c^{\ast} \vert= \vert\Omega_c \vert$, it might be helpful. Then it remains to show:
$$\vert \lbrace x\in\Omega^{\ast}: f^{\ast}(x)\geq c\rbrace \vert = \vert \lbrace x\in\Omega: f(x)\geq c\rbrace ^{\ast} \vert$$
Are both sets within $\vert\cdots \vert$ in the last equation the same? That is, do we have:
$$\lbrace x\in\Omega^{\ast}: f^{\ast}(x)\geq c\rbrace = \lbrace x\in\Omega: f(x)\geq c\rbrace ^{\ast}$$
If the last statement is true, then 1) and 2) becomes trivial. But I do not know how to prove any of the three statements.  
 A: Yes, the equality
$$ \lbrace x\in\Omega^{\ast}: f^{\ast}(x)\geq c\rbrace  =   \lbrace x\in\Omega: f(x)\geq c\rbrace ^{\ast}  $$
is true, which implies affirmative answers to your other questions. Note that the set on the right is $\Omega_c^*$. So the claim is that 
$$
f^*(x)\ge c \iff x\in \Omega_c^* \tag1
$$ 
The implication $\impliedby$ is immediate because $f^*(x):=\sup\lbrace{ c\geq 0:x\in \Omega_c^{\ast} \rbrace}$. 
Conversely, if $f^*(x)\ge c$ then for every $n\in\mathbb{N}$ it holds  that $x\in \Omega_{c-1/n}^{\ast}$. Note that
$$
\Omega_c = \bigcap_{n\in\mathbb{N}} \Omega_{c-1/n}
$$
which by the continuity of measure implies (this is where we need the measure to be finite, which it is because of boundedness):
$$
\left|\Omega_c\right| = \inf_{n\in\mathbb{N}}|\Omega_{c-1/n}|
$$
Since the measure determines the  Schwarz symmetrization of these sets, we have
$$
\Omega_c^* = \bigcap_{n\in\mathbb{N}} \Omega_{c-1/n}^*
$$
(the intersection of closed balls of radii $r_n\searrow r$ is the closed ball of radius $r$).
Thus $x\in \Omega_x^*$, proving (1).
