Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.