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I came upon this question here which contains the following statement:

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.

I believe that Lipschitz here refers to Lipschitz continuity.

When I saw the statement it seemed to me that Lipschitz continuity was too strong. For example, Lipschitz continuous implies continuous. Is it conceivable that a continuous (read "nice") function maps a measurable set to a non measurable set?

It seems to me that this would be the only case for which the image can have non zero measure since it seems intuitively imperative that any image cannot have larger measure than the original set.

Please could someone enlighten me on the minimal condition for which $f: \mathbb R^n \to \mathbb R^m$ maps null sets to null sets? And if possible point out any mistakes in my thoughts above, I would greatly appreciate it.

Note: I expect the answer to be that $f$ should be measurable or some similarly weak condition.

Community
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a student
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    What you are looking for is absolute continuity. The Cantor function is the standard example of a uniformly continuous function that does not map null sets to null sets. – Prahlad Vaidyanathan Jun 20 '15 at 06:13

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No, there is a continuous function from the unit interval that sends the Cantor set onto the unit interval. Since the Cantor set is null, the contradicts your conjecture.

See https://en.m.wikipedia.org/wiki/Cantor_function

Since all continuous functions are measurable, your guess after the question is also wrong.

Absolute continuity is sufficient, but be clear- there are plenty of discontinuous functions that have this property. It will be unlikely that you'll find a minimal condition in any strict notion of "minimal." And it's not entirely clear how to extend the definition of absolute continuity to multiple dimensions.

Thomas Andrews
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  • But doesn't absolute continuity imply Lipschitz continuity? In this case, the condition you suggest is even stronger than the condition in the linked question... – a student Jun 20 '15 at 07:12
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    @a student: No, Lipschitz continuity implies absolute continuity (in dimension one at least). Further, it might be interesting to you that a (continuous) increasing function is absolutely continuous iff it maps null sets to null sets iff it maps measurable sets to measurable sets. Generally, mapping null sets to null sets and measurable sets to measurable sets is equivalent for continuous maps. – PhoemueX Jun 20 '15 at 07:17
  • No, there is no reason to think absolute continuity implies Lipshitz, not that I know of. Lipschitz clear implies absolute, and Lipshitz is not listed as one of the equivalences of absolute continuity. @astudent – Thomas Andrews Jun 20 '15 at 12:02