Hausdorff measure vs Lebesgue measure for a hypersurface in $\mathbb{R}^n$

Question

Let $H$ be a compact smooth hypersurface with boundary in $\mathbb{R}^n$. We can compute the Lebesgue measure $\mathcal{L}(H)$ with respect to the induced Lebesgue measure coming from $\mathbb{R}^n$, and also separately the Hausdorff measure $\mathcal{H}^{n - 1}(H)$. My question is, how do they compare? Does one always have $\mathcal{L}(H) \leq \mathcal{H}^{n - 1}(H)$?

Answer 1

A lower dimensional surface always has Lebesgue measure zero. Indeed, if $j < k$, and $\mathcal H^j(E) < \infty$, $\mathcal H^k(E)=0$. This is a straightforward consequence of the definition of the Hausdorff measure; if this isn't obvious, you should try proving it yourself. On the other hand $\mathcal H^n = \mathcal L$. It's not so hard to see that $\mathcal H^n \leq \mathcal L^n$, but the reverse inequality is harder. However, it isn't hard to see that $\mathcal L^n \leq C_n \mathcal H^n$ for some constant $C_n$, and this is enough to show you that $\mathcal L(M)=0$ when $M$ has finite $\mathcal H^{n-1}$ measure. To see this inequality, take a competing cover in the definition of the Hausdorff measure, and show that the covering sets can be replaced by cubes with comparable diameter.

I suspect what you meant, though, is if the Hausdorff measure was equal to some "natural" lower dimensional measure on a hypersurface. One can define a lower dimensional measure on a surface without going through the Hausdorff measure. This is basically the same construction you use in multivariable calculus classes, where you compute the measure of a parametrized surface by looking at the areas of the images of infinitesimal paralellograms. First, for a smooth, injective map $f : \mathbb R^{n-1} \to \mathbb R^n$, define $\mu_f(E) = \int_{f^{-1}(E)} \det(D_x f^T D_x f) d \mathcal L^{n-1}(x)$. If $M \subset \mathbb R^{n}$ is a smooth hypersurface, then there are finitely $\phi_i:\mathbb R^{n-1} \to M$ which are diffeomorphisms, such that $M = \bigcup_i Im(\phi_i)$. Let $U_i = Im(\phi_i)$, and let $\rho_i$ be a partition of unity subordinate to this open cover of $M$. Define $\mu_M = \sum_i \rho_i \mu_{f_i}$.

There is a result called the area formula in geometric measure theory which says that the $\mu_M$ defined above and $\mathcal H^{n-1}$ restricted to $M$ are equal. This will be proved in most books on GMT, in particular there is a nice proof of it in Francesco Maggi's book (note: the equality $\mathcal H^n = \mathcal L^n$ alluded to earlier is also proved in this book). You might ask if the area formula characterizes $\mathcal H^{n-1}$ in some reasonable sense. Anything worthy of being called a lower-dimensional measure better satisfy the area formula, but are there other measures which are different from $\mathcal H^{n-1}$ which also satisfy the area formula? The answer turns out to be yes. For example, the Favard measures, also called integralgeometric measures, also satisfy the area formula, but they differ wildly from the Hausdorff measure on unrectifiable sets, i.e. sets which are not Borel subsets of countable unions of $C^1$ manifolds.