Hausdorff measure of an embedded submanifold

Let $M$ be a $C^1$- embedded n-submanifold (without boundary) of $R^{n+k}$. Is it true that for every $K$ compact set in $R^{n+k}$ the n-dimensional Haussdorf measure of $M \cap K$ is finite?

Thanks

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I don't think so... perhaps a 'very long' (infinitely long) zigzaging bounded curve in $\mathbb{R}^2$ would do the trick? Like a smooth version of Koch's snowflake or something. I'm just guessing here, I don't know if I'm actually right ; I don't know if a bounded closed curve with infinite length can be $C^1$. If you only assume $C_0$ though, Koch's snowflake is a counter-example. –  Patrick Da Silva Jun 5 '13 at 10:38
@PatrickDaSilva: I think, you shall think of $\sin \frac1x$-like curves as a subset of $[0,1]^2$. –  Ilya Jun 5 '13 at 11:34

Not in general, as the example $y=\sin(1/x)$, $0<x<1$, shows (given by Ilya as a comment).
But the answer is yes if $M$ is a closed subset of $\mathbb R^{n+k}$. Indeed, by the usual compactness argument it suffices to show that every point $x\in \mathbb R^{n+k}$ has a neighborhood $U$ such that $\mathcal H^n(M\cap U)<\infty$.
• If $x\notin M$ then there is a neighborhood $U$ such that $M\cap U=\varnothing$.
• If $x\in M$ then there is a neighborhood $V$ such that $M\cap V$ is diffeomorphic to a hyperplane with a diffeomorphism of $V$. This diffeomorphism is Lipschitz on any compactly contained neighborhood $U\Subset V$. This $U$ satisfies $\mathcal H^n(M\cap U)<\infty$.