Assume bounded domain (open and connected) $\Omega\subset \mathbb{C}^n$, and a smooth function $\rho:\mathbb{C}^n\longrightarrow \mathbb{R}$ such that $\rho(x) = 0$ for all $x\in \partial \Omega$, $\rho (x) < 0$ for all $x\in \Omega$ and $\rho(x) > 0$ for all $x\in \mathbb{C}^n\backslash \overline{\Omega}$. And $\nabla \rho \neq 0$ on $\partial \Omega$. We say $\partial \Omega$ is smooth. Now let $z\in \partial \Omega$, we denote by $\tau(z)$ the expression $$ \tau(z) = \lim_{\varepsilon \longrightarrow 0^+} \frac{vol(S(z,\varepsilon)\cap \Omega)}{vol(S(z,\varepsilon))}$$ where $S(z,\varepsilon) = \{x\in \mathbb{C}^n: |x - z| = \varepsilon\}$ -- the boundary of ball $B(z,\varepsilon)$ in $\mathbb{C}^n$ as usual. And $vol$ denote the usual measure on $\partial \Omega$.
Show that when $\rho$ is smooth as above, we always have $\tau(z) = \frac{1}{2}$ for all $z\in \partial \Omega$.
Furthermore, if $\partial \Omega$ is piece-wise smooth, then $\tau(z)$ is defined on $\partial \Omega$ and not zero.
I read this result in the book "The Bochner--Martinelli Intergral and It's application" but I don't understand why this fact is trivial - as the author said.
This has northing to do with complex analysis because there is nothing about a holomorphic function anywhere. So it must be a fact just about embedded submanifolds of $\mathbb{R}^{n}$.
Intuitively, as $\epsilon \to 0$, the surface cutting the ball approaches its tangent plane. The tangent plane obviously cuts the ball in half.
So what is left it to use estimate the difference between the two volumes, for which I would imagine Taylor's theorem would come in handy (actually just the definition of the derivative I think).
Try this and if you get stuck I will see about adding more details.