Measure of level sets If there's a bounded domain $\Omega\subset\mathbb{R}^n$ and a function $u\in L^1(\Omega)$, why does
$\mathcal{L}^n\{\Omega \cap x:u(x)=t \}=0$ hold für all $t\in\mathbb{R}$ except for $t\in S$ with $S\subset \mathbb{R}$ a countable subset?
 A: Let $S = \{t\ |\ \mu(u^{-1}(t)) > 0\}$ and $S(a, b) = \{t | |t| > a \wedge \mu(u^{-1}) > b\}$.
Note that $S\setminus \{0\} = \bigcup\limits_{n,m} S\left(\frac{1}{n}, \frac{1}{m}\right)$. If each $S\left(\frac{1}{n}, \frac{1}{m}\right)$ is countable then $S$ is countable as countable union of countable sets, so if $S$ is uncountable then $S\left(\frac{1}{n_0}, \frac{1}{m_0}\right)$ is uncountable for some $n_0$ and $m_0$.
Now we have $$\int_\Omega |u(x)|\, d\mu(x) \geq\\
\int_{u^{-1}\left(S\left(\frac{1}{n_0}, \frac{1}{m_0}\right)\right)} |u(x)|\,d\mu(x) \geq \\
\sum\limits_{t \in S\left(\frac{1}{n_0}, \frac{1}{m_0}\right)} \ \int\limits_{u^{-1}(t)} |u(x)|\, d\mu =\\
\sum\limits_{t \in S\left(\frac{1}{n_0}, \frac{1}{m_0}\right)} \ \int\limits_{u^{-1}(t)} |t|\, d\mu =\\
\sum\limits_{t \in S\left(\frac{1}{n_0}, \frac{1}{m_0}\right)} |t| \cdot \mu(u^{-1}(t)) \geq\\
\sum\limits_{t \in S\left(\frac{1}{n_0}, \frac{1}{m_0}\right)} \frac{1}{n_0 \cdot m_0} = \infty
$$
The last sum is infinite because we have infinitely many terms separated from $0$.
So we have $u \notin L^1(\Omega)$ - contradiction, thus $S$ is countable.
