A level set of non-constant real analytic function Assume that $u: B\to [0,1]$, where $B$ is an open ball in $\mathbb{R}^n$, is a nonconstant real analytic function and let $t\in[0,1]$ and let $\mu(t)=\operatorname{Volume}(\{x\in B: u(x)>t\})$. Why is $\mu$ continuous? I need a reference.
 A: Yes. The function $\mu$ is continuous. The way to show this is by turning this problem into the question of whether the zero sets of analytic function are of measure zero and then solving this problem.
Observation (standard fact from measure theory): Let $(X,\mathcal{M},m)$ be a measure space where $\mu(X)<\infty$, and let $u:X\rightarrow\mathbb{R}$ be a measurable function. Then the function
$\mu:\mathbb{R}\rightarrow\mathbb{R}$ defined by $\mu(t)=m(\{x\in X\mid u(x)>t\})$ is continuous if and only if $m(\{x\in X\mid u(x)=t\})=0$ for all real numbers $t$.
Suppose that $U\subseteq\mathbb{R}^n$ is open and connected and $f:U\rightarrow\mathbb{R}$ is analytic and non-constant. I claim that the zero set $Z(f)=\{x\in U:f(x)=0\}$ has measure zero. To see this, we observe that if $x\in U$, then some higher order derivative of $f$ at the point $x$ is non-zero; if all the derivatives of $f$ were zero, then we would have $f(x)=0$ everywhere, and we have excluded that case. To continue the proof, we will use the following fact about differentiable functions.
If $U$ is an open subset of $\mathbb{R}^n$ and $f:U\rightarrow\mathbb{R}$ is differentiable and has non-zero gradient everywhere, then $Z(f)$ is a differentiable sub manifold of $U$. Since submanifolds of $U$ have measure zero, the zero set $Z(f)$ also has measure zero.
I now claim that each $x\in U$ has a neighborhood $V$ with $x\in V\subseteq U$ where $Z(f)\cap V$ has measure zero. If $x\in U$, then some $n$-th order derivative $\frac{\partial^n}{\partial x_{1}^{a_1}\dots\partial x_{1}^{a_r}}f$ is non-zero, so we will prove that there exists such a neighborhood $V$ by induction on $n$. Let $y\in U$, and let $V$ be a neighborhood of $U$ such that if $z\in V$, then $\frac{\partial^n}{\partial x_{1}^{a_1}\dots\partial x_{1}^{a_r}}f(z)\neq 0$. If $n=0$, then our proof is complete, so assume that $n>0$.
Then $\frac{\partial^{n-1}}{\partial x_{1}^{a_1}\dots\partial x_{1}^{a_r-1}}f$ has a non-zero gradient on $V$, so $Z(\frac{\partial^{n-1}}{\partial x_{1}^{a_1}\dots\partial x_{1}^{a_r-1}}f)\cap V$ has measure zero. On the other hand, by the induction hypothesis, and using the fact that every open subset of $U$ is Lindelof, we know that $Z(f)\cap V\setminus Z(\frac{\partial^{n-1}}{\partial x_{1}^{a_1}\dots\partial x_{1}^{a_r-1}}f)$ has measure zero as well. Therefore, we know that $$Z(f)\cap V\subseteq \big(Z(f)\cap V\setminus Z(\frac{\partial^{n-1}}{\partial x_{1}^{a_1}\dots\partial x_{1}^{a_r-1}}f)\big)\cup\big(Z(\frac{\partial^{n-1}}{\partial x_{1}^{a_1}\dots\partial x_{1}^{a_r-1}}f)\cap V\big),$$ so $Z(f)\cap V$ has measure zero since it is the subset of the union of two measure zero sets.
