# Surface measure for Lipschitz domain

Let $D\subset R^d$ be a bounded Lipschitz domain. Must there exist a bounded function $\Phi$ on $\partial D$ and collections of subsets $(\partial D )^{\epsilon} \subset \partial D$ (indexed by $\epsilon$) such that for all bounded continuous function $g$ on $\partial D$ we have $$\lim_{\epsilon\to 0} \sum_{(\partial D )^{\epsilon}}g\,\Phi\,\epsilon^{d-1}= \int_{\partial D }g\,d\sigma$$

We may suppose $\partial D$ is just the graph of a Lipschitz function over a ball in $R^{d-1}$.

This is a discrete approximation to the surface measure $\sigma$. The result is true if we add the condition that "the unit outward normal vector field on $\partial D$ is continuous $\sigma-$almost everywhere on $\partial D$." But I wonder if we can get rid of it.

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Let $\sigma$ be the restriction of the $(d-1)$-dimensional Hausdorff measure to $\partial D$. Since $D$ is bounded with Lipschitz boundary, $\sigma$ is a finite measure. The Krein-Milman theorem implies that $\sigma$ is a weak*-limit of sums of point masses: that is, there exists a sequence $\nu_k$ of finite combinations of point masses (with nonnegative coefficients and total mass equal to $\sigma(\mathbb R^d)$ ) such that $\int g\,d\nu_k\to \int g\,d\sigma$ for every function $g\in C(\mathbb R^d)$. This yields the desired convergence.
Putting the measures $\nu_k$ into the desired form, even with $\Phi\equiv 1$, should be straightforward: take the mass of each point to be a rational number, hide their common denominator in $\epsilon^{d-1}$, and split integer masses into tightly bunched unit masses.