The $\mathcal H^{N-1}$ measure of the projection of a curve is always less then itself

Question

Let $\Gamma \subset \mathbb R^N$ be a $\mathcal H^{N-1}$-rectifiable set such that $\mathcal H^{N-1}(\Gamma)<\infty$.

Let $P(\Gamma)$ be the projection of $\Gamma$ onto the hyperplane $\{x_N=0\}$. I am trying to prove that $$ \mathcal H^{N-1}(P(\Gamma))\leq \mathcal H^{N-1}(\Gamma) $$ I think it is true, at least for some easy geometry it is easy to see... But I don't know how to prove it in general...

Any help is really welcome!

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PS: I don't think rectifiable is useful here. But I have it anyway...

Answer 1

This follows from the following lemma:

If $f$ is a Lipschitz map with Lipschitz constant $L$, then $\mathcal{H}^s(fE) \le L^s \mathcal{H}^s(E)$ for all $s$.

This follows in a couple lines from the definition of Hausdorff measure: Take a cover $\{E_i\}$ of $E$ and notice that $fE_i$ is a cover and $$\operatorname{diam}(fE_i) \le L \operatorname{diam} E_i$$

For a reference, this is Mattila chapter 7 or somewhere in Falconer.