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Follow up to this question:

Suppose we define function $f:[0,1]\to [0,1]$ that is measurable in the Caratheodory sense, using the Lebesgue Outer Measure, and the graph of $f$ is dense in $[0,1]\times[0,1]$.

Main Question:

Can the pre-image of a non-measurable subset of $[0,1]$ (under $f$) be measurable in the sense of caratheodory, using the Lebesgue Outer Measure?

Attempt:

I'm not sure how to approach the main question. I attempted a solution by taking the proposed contrapositive of the main question:

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

but this isn't the contrapositive as stated from @S.L.'s comment to the original question.

Furthermore, due to my lack of formal training beyond Intro to Advanced Mathematics, I don't know the proper answer to the main question. (However, I do know non-measurable sets can't be explicitly defined and vary by their Lebesgue Outer Measure.)

Even then, I'm not sure if this means the pre-image of a non-measurable set under $f$ (even for continuous, non-constant $f$) can never be measurable, proving the main question wrong.

Final Note/Motivation:

I wanted to make sure the main question in this post (specifically criteria 3.) to help satisfy the motivation of the 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 $[0,1]\times[0,1]$.

If the main question for this post is wrong, I would like someone to rewrite the main question in this post to satisfy the motivation.

Arbuja
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The same function given by https://math.stackexchange.com/a/4706564/1184648 also works as an example here. Take a Vitali set $V$ with $V \cap \mathbb{Q}=\{1\}$. Then the function $f$ defined as in the answer to that post satisfies $f^{-1}(V)=\emptyset$.

FZan
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