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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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    Regarding your first sentence: is there anything to prove? If $f$ is a function, the cardinality of the image will be less equals the cardinality of $A$, so will have smaller equal Lebesgue measure. In particular: why do you require $f$ to be Lipschitz?2012-09-14
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    It seems like you are looking for Lusin N-property.2012-09-14
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    @Matt The cardinality argument argument tells you nothing and is in fact wrong. See [here](http://math.stackexchange.com/q/90225/5363) for a counterexample (the image of the Cantor set under the Cantor-Lebesgue function has measure one) and [here](http://math.stackexchange.com/q/139883/5363) for a proof of the statement in the question.2012-09-14
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    Matt: Let $C$ be the Cantor set and $f$ the [function](http://en.wikipedia.org/wiki/Cantor_set#Cardinality) that maps $C$ onto $[0,1]$. Isn't that a counterexample to your claim?2012-09-14
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    @Matt: In $\mathbb R^2$, the cardinality of the unit disk (Lebesgue measure $\pi$) and the cardinality of a straight line (Lebesgue measure $0$) is the same. Thus there exists a surjective function from the straight line to the unit disk. Of course you can extend that function to all of $\mathbb R^2$.2012-09-14
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    @celtschk: and you can make it continuous by taking a Peano curve and using Tietze's extension theorem...2012-09-14
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    @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.2012-09-14
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    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.2012-09-14
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    @t.b. Thank you lots for the comment. I thought it might be wrong and hoped someone would tell me if I commented. I will read the links, but right now I'm too tired.2012-09-14
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    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." :-)2012-09-15
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    I don't understand how you can be looking for a condition which doesn't involve the metric structure. If the two measures are arbitrary rather than depending on the metric space (eg Hausdorff measure), then nothing about the map itself can tell you what sets will be nullsets in the image measure space (for instance, take some measure on the image space, and add a measure supported on the image of some nullset).2013-12-22
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    @jwg Good point (it seems rather obvious now); if you post your comment as an answer I will accept it.2013-12-22

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There is no real condition on a map which could be valid for arbitrary measure spaces. If the measures can be arbitrary, rather than depending on the metric space structure of the underlying space, (eg Hausdorff measure), then nothing about the map itself can tell you which sets will be nullsets in the image measure space.

For example, given some measure on the image space and a map from some other measure space which sends nullsets to nullsets, add a measure supported on the image of some nullset. The map no longer maps nullsets to nullsets.