Hausdorff Dimension of a Smooth Manifold

Question

I read a book about fractals which stated without proof:

Every $m$-dimensional $(m<n)$ smooth manifold $M$ in $\mathbb{R}^n$ has Hausdorff dimension $m$.

How can we prove this?

Answer 1

Let $d < m < e$. It suffices to show $0 = \mathcal{H}^d(U) < \mathcal{H}^e(U) = \infty$ for one smooth chart $(U,\phi)$ around each point $p \in M$, since $M$ can be covered with countably many of these ($M$ is Lindelöf!). But for each point we can find some small neighborhood $U$ which can be written as graph of a smooth function having its domain in an $m$-dimensional affine subspace of $\mathbb{R}^n$ and mapping into the orthogonal complement of that subspace. But such graphs have finite $m$-dimensional Hausdorff-measure as Lipschitz-transformations of bounded subsets of $m$-dimensional affine spaces.