Let $S$ be a sufficiently regular surface in $\mathbb{R}^d$ (say parametrized by a smooth function $f : [0,1]^{d-1} \rightarrow \mathbb{R}^d$). There are two natural definitions of its surface area:
- The $d-1$ dimensional Hausdorff measure $\mathcal{H}^{d-1}(S)$.
- $\text{Area}(S) := \lim_{\epsilon \rightarrow 0} \frac{\mathcal{H}^d(S_{\epsilon})}{2\epsilon}$, where $S_{\epsilon}$ denotes the $\epsilon$-inflation $\{x \in \mathbb{R}^d \mid d(x,S) < \epsilon\}$ of $S$.
Question: Is there a simple way to prove that these definitions are equivalent?
In the case $S$ is the image of a smooth injective function $f : [0,1]^{d-1} \rightarrow \mathbb{R}^d$, the equality can be obtained by considering the function $F : [0,1]^{d-1} \times \mathbb{R} \rightarrow \mathbb{R}^d : (x, t) \mapsto f(x) + t \cdot v(x)$, where $v$ is a vector field of unit norm orthogonal to $S$. However, this method is fairly tedious and doesn't easily generalize to less regular sets $S$.
I am wondering whether there is a more direct proof of the equivalence, that does not resort to mathematical structures other than the Hausdorff measures. Of course, some regularity assumptions are needed for the equivalence to hold. I don't know enough about geometric measure theory to figure it out.