Let $D$ be a smooth domain of $\mathbb{R}^d$. Let $\partial D$ denote the boundary of $D$. We denote by $B(x,r)=\{y \in \mathbb{R}^d \mid |y-x|<r\}$ the Euclidean ball centered at $x$ with radius $r$


For $x \in \partial D$, we consider the following limit:
\begin{align\*}
\lim\_{r \to 0}\frac{1}{m(D \cap B(x,r))}\int\limits\_{D \cap B(x,r)}\frac{z-x}{r}\,m(dz)
\end{align\*}
Here, $m$ denotes the Lebesgue measure on $\mathbb{R}^d$.


Does this limit exists?


In my impression, this seems to converge to the inward (unit) normal vector at $x$.



On a characterization of inward unit normal vector
For small $r$ the curvature of the surface $\partial D$ can be neglected, so $D \cap B(x,r)$ is half the $d$-dimensional ball with radius $r$. Choosing the origin of the coordinate system at position $x$ and orienting the $x\_1$-axis along the inward normal, the integral is given by the vector $v$ with components
$$v\_p=\frac{2}{V\_{d}(r)}\int\cdots\int\_{-\infty}^\infty \theta\biggl(r-\sum\_{i=1}^d x\_i^2\biggr)\theta(x\_1)\frac{x\_p}{r}\,dx\_1dx\_2\ldots dx\_d$$
with $V\_d(r)$ the volume of the $d$-dimensional ball of radius $r$ and $\theta$ the unit step function.  

Only the $p=1$ component is nonzero because of symmetry, and this component evaluates to
$$v\_1=\frac{\int\_0^r d\rho\int\_0^{\pi/2}d\theta\,\rho^{d} \cos \theta \sin^{d-2}\theta}{r\int\_0^r d\rho\int\_0^{\pi/2}d\theta\,\rho^{d-1} \sin^{d-2}\theta}=\frac{\Gamma \left(\frac{d}{2}+1\right)}{\sqrt{\pi }\, \Gamma \left(\frac{d+3}{2}\right)}.$$
So it's an inward normal vector, but not a unit vector.


