When is the image of a null set a null set? 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.
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.
Do you know Peano curve ?

It maps $[0,1]$ to an area in $\mathbb{R}^2$ with positive 2 dimensional Lebesgue measure