Volume form and Hausdorff measure

Question

Let $M$ be a smooth orientable $(n-1)$-dimensional submanifold in $\mathbb{R}^n$, $dS$ be its volume form and $dH^{n-1}(x)$ be an $(n-1)$-dimensional Hausdorff measure. How to show than that $$ \int\limits_{M} f(x) dS = \int\limits_{M} f(x) dH^{n-1}(x) $$

In fact, it is a generaliation of an equality formula for surface integrals of first and second kind in $\mathbb{R}^3$.

Answer 1

Assume that $M = \{x \in \mathbb{R}^n \mid g(x) = 0 \}$ and $\partial_1 g > 0$. Then by the area formula we have $$ I = \int\limits_{M}f(x)dH^{n-1}(dx) = \int\limits_{\mathbb{R}^{n-1}}f(y) J_[\varphi](y) dL^{n-1}(y) $$ where $J_{n-1}[\varphi]$ is a $(n-1)$-dimensional Jacobian of the mapping $\varphi(y_1,...,y_{n-1}) = (x_1(y),y_1,...,y_{n-1})$, where $x_1(y)$ is such that $g(x_1(y),y_1,...,y_{n-1})=0$. So $J_{n-1}(y) = \frac{|\nabla g|}{\partial_1 g}$. By the other hand $\frac{|\nabla g|}{\partial_1 g} y_1 \wedge ... \wedge y_{n-1}$ is a pullback of the form $\omega$ such that $\frac{dg}{| \nabla g |} \wedge \omega = \mathrm{dx}$. The volume form $dS$ on $M$ satisfies the same equation. Then $$ I = \int\limits_{M} f(x)dS $$ If $M$ is arbitrary we must use a partition of unity.

Answer 2

You can use the area formula. The following argument appears as section 3.3.4, example D in Evans and Gariepy's Measure Theory and Fine Properties of Functions. Let $f:V\to M$ for $V \subset \mathbb{R}^{n-1}$ be a coordinate map, then the area formula (section 3.3.2, theorem 1) says that for $\eta\in C^\infty(M)$, $\int_V \eta Jf \,dLeb_{n-1}=\int_{\mathbb{R}^n} \sum_{x \in f^{-1}(\{y\})} \eta(x) \, d\mathscr{H}^{n-1}(y)=\int_M \eta \, dS$. But writing $g_{ij}=\frac{\partial f}{\partial x^i} \cdot \frac{\partial f}{\partial x^j}$ for the Euclidean inner product $\cdot$, we have $(Df)^* \circ Df=((g_{ij}))$, so $Jf=\sqrt{\det(g)}$. By the standard result (e.g. Prop. 15.31 in Lee's Introduction to Smooth Manifolds), we have $\int_V\sqrt{\det(g)} \, dLeb_{n-1}=\int_M\,dS$.