[Math] 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.

Best Answer

Just posting this in case others search for it later. As @Leonid mentioned, here is the theorem from Geometry of sets and measures in Euclidean spaces by Pertti Mattila:

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, $$\dim(f(A)) \leq \dim(A).$$

Related Question