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

A: The quoted (expensive) book doesn't even supply a proof!  The case with Hausdorff Measure is much more complicated, as well.
Assume known that a Lipschitz function sends null sets to null sets.  Let $A$ be measurable, and approximate it from the outside by a countable p.w.-disjoint union of open balls $B = \uplus B_j$, so that $\mu(B \setminus A) < \epsilon$, which we can do by the construction of the Lebesgue measure.
Then $\mu(f(A)) \leq \mu(f(B)) \leq \sum \mu(f(B_i)) \leq \sum L \cdot \mu(B_i) = L \cdot \sum \mu(B_i) = L \cdot \mu(B)$
Now let $\mathcal{B}$ be a sequence of such $B$'s so that $\epsilon \rightarrow 0$ and the conclusion follows.  You can see that it's "not even close", per se, with many points at which a strict inequality could arise.
