Measure of image of Lipschitz function is bounded?

I recently watched some measure theory lectures online. They didn't post lecture notes and I can't find which video exactly it was.

I think there was a theorem that goes something along the lines of:

If $f:\mathbb{R^N} \to \mathbb{R^N}$ is Lipshitz with Lipschitz constant $L$, and $\lambda$ stands for Lebesgue measure, then $\lambda(f(A)) \leq L\lambda(A)$ for $A$ measurable.

Is this correct, or is there a similar looking theorem that I might be thinking of? Thanks.

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Yes, it's correct, although the proof I know uses Hausdorff measure. – Jose27 Jun 14 '12 at 1:18
Does it have a name, or do you have a link to it? The proof I remember involved Hausdorff measure too. – nullUser Jun 14 '12 at 1:25
I don't have a reference, although, when $\lambda(A)=0$ this is called Luzin's N property for Lipschitz maps. – Jose27 Jun 14 '12 at 1:34
Theorem 7.5 in Geometry of sets and measures in Euclidean spaces by Pertti Mattila, a highly recommended book for everyone who is interested in anything that involves maps and measures. – user31373 Jun 14 '12 at 2:27

7.5. Theorem. If $f:\mathbb{R}^m \to \mathbb{R}^n$ is a Lipschitz map, $0 \leq s \leq m$, and $A \subset \mathbb{R}^m$, then $$\mathcal{H}^s(f(A)) \leq \mathrm{Lip}(f)^s\mathcal{H}^s(A).$$ In particular, $$\dim(f(A)) \leq \dim(A).$$