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Here is the original question:

Suppose we define function $f$ that is measurable in the Caratheodory sense.

Main Question:

Can the pre-image of a non-measurable set under $f$ be measurable in the sense of caratheodory?

Attempt:

I'm not sure how to approach the main question but we will attempt a solution by taking the contrapositive of this question:

Can the image of a measurable set under $f$ be non-measurable?

Due to my lack of formal training beyond Intro to Advanced Math, I couldn't figure out if the contrapositive is true; however, from Royden in Real Analysis, if the Cantor-Lebesgue function is $f$, there exists a measurable subset in the domain of $f$ (i.e. the cantor set) whose image under $f$ is non-measurable.

Is the second question infact the contrapositive of the main question? If not, how do we prove the main question is wrong?

Final Note/Motivation:

I wanted to make sure the main question in this post (specifically criteria 3.) helps to satisfy the motivation of that post.

In other words, I want to find a function $f:[0,1]\to[0,1]$ whose graph is dense, and somewhat but not too evenly distributed, in $[0,1]\times[0,1]$

Sarvesh Ravichandran Iyer
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Arbuja
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    If the image of a measurable set is non-measurable, that doesn't mean that the preimage of the non-measurable set is measurable, because there may have been other sets that map into our non-measurable set. – S.L. May 24 '23 at 18:23
  • @S.L. I see, but is my main question true? – Arbuja May 24 '23 at 18:27
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    a silly answer to the question in the title would be to take a constant function and the preimage under a non-measurable set containing the single point in the image... (which will then just be the domain, measurable) – George C May 24 '23 at 18:27
  • @GeorgeCoote Thanks. I wanted to make sure that criteria 3. in the main question of this post helps satisfy the motivation of the post. – Arbuja May 24 '23 at 18:36
  • @GeorgeCoote I don't find this answer to be silly at all :) – Maximilian Janisch May 25 '23 at 08:24

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