show/hide this revision's text 2 Adding nothingness

My guess is that the characterization is the following:

A function $f$ maps every non-measurable set into a measurable set if and only if the domain or the image of $f$ has measure zero.

One direction is trivial. For the other direction assume that the image of $f$ is positive. Take a non-measurable subset $N$ of the image and a measurable subset $M$ of the image so that

  1. $N$ and $M$ are well separated.
  2. $f^{-1}(N)$ and $f^{-1}(M)$ are well separated.
  3. $f^{-1}(M)$ has positive measure.

Take a non-measurable subset $K$ of $f^{-1}(M)$ and consider $K \cup f^{-1}(N)$. This set is non-measurable and so is its image under $f$.

Is

Are there a mistake more mistakes hidden somewherein this idea?
show/hide this revision's text 1

My guess is that the characterization is the following:

A function $f$ maps every non-measurable set into a measurable set if and only if the image of $f$ has measure zero.

One direction is trivial. For the other direction assume that the image of $f$ is positive. Take a non-measurable subset $N$ of the image and a measurable subset $M$ of the image so that

  1. $N$ and $M$ are well separated.
  2. $f^{-1}(N)$ and $f^{-1}(M)$ are well separated.
  3. $f^{-1}(M)$ has positive measure.

Take a non-measurable subset $K$ of $f^{-1}(M)$ and consider $K \cup f^{-1}(N)$. This set is non-measurable and so is its image under $f$.

Is there a mistake hidden somewhere in this idea?