# When is the image of a null set also null?

It is easy to prove that if $A \subset \mathbb{R}$ is null (has measure zero) and $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lipschitz then $f(A)$ is null. You can generalize this to $\mathbb{R}^n$ without difficulty.

Given a function $f: X \rightarrow Y$ between measure spaces, what are the minimal conditions (or additional structure) needed on $X$, $Y$ and $f$ for the image of a null set to be null?

Any generalization (containing the above as a special case) is appreciated. Apparently if $X$ and $Y$ are $\sigma$-compact metric spaces with the $d$-dimensional Hausdorff measure and $f$ is locally Lipschitz then the result holds. Can we be more general? I would like to see something without a metric.

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It seems like you are looking for Lusin N-property. – Nikita Evseev Sep 14 '12 at 10:25
@Matt The cardinality argument argument tells you nothing and is in fact wrong. See here for a counterexample (the image of the Cantor set under the Cantor-Lebesgue function has measure one) and here for a proof of the statement in the question. – t.b. Sep 14 '12 at 10:39
@nikita2: That property references the real line and Lebesgue measure. I'm basically asking for sufficient conditions for a generalization of the Lusin N property. – Ricardo Buring Sep 14 '12 at 11:12
There are results for Sobolev maps (more general than the Lipschitz maps), but here the metric structure is even more involved than in the Lipschitz case. I never saw a nontrivial condition for property N which did not involve a metric. A trivial non-metric condition would be: all subsets of $Y$ with positive measure have larger cardinality than any subset of $X$ with zero measure. – user31373 Sep 14 '12 at 15:31
Of course the minimal condition can be formulated quite easily: It is "the function has the property that it maps null sets to null sets." :-) – celtschk Sep 15 '12 at 14:19