If $\mu\{\partial \{u\ge t\}\}<\infty$ for all $t>0$ then $\mu(\partial\{u\ge t\})>0$ for at most countably many $t$?

Question

$u \in C_c(X)$. Let $t>0$ and $\partial \{u\ge t\} \subset \{u=t\}\subset supp u$ and $\mu(supp u)<\infty$. Then why is $\mu(\partial\{u\ge t\})>0$ for at most countably many $t$?

Answer 1

Set $L_t = \{u\ge t\}$. By continuity, $\partial L_t= \{u=t\}$. By definition, the support of $u$ is the closure of $\{|u|>0\}$, and by assumption, has finite $\mu$-measure.

For $t \ne t'$, we have that $\partial L_t \cap \partial L_{t'}=\emptyset$.

If the set $A= \{t>0:\mu(\partial L_t)>0\}$ is uncountable, there exists some $n_0$ such that for any $n>n_0$, there exist infinitely many elements $t\in A$ with the property $\mu(\partial L_t)> \frac 1n$ (why?), and it follows that we can find a countable subset $\{t_1,t_2,\dots\}$ of $A$, such that $\mu(\partial L_{t_j}) > \frac{1}{j}$ for all $j$ large enough. By $\sigma$-additivity,

$$ \mu (\cup_{j=1}^\infty \partial L_{t_j}) = \sum_{j} \mu (\partial L_{t_j}) = \infty.$$

However, for any $t\in A$, $\partial L_t$ is contained in the support of $u$, and therefore so is the union $\cup_{j=1}^\infty \partial L_{t_j}$, since the support has finite measure, also $\mu (\cup_{j=1}^\infty \partial L_{t_j})<\infty$, a contradiction. Thus $A$ is countable.