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This is a follow up to this post. Note in the main question of this post, $f$ can have a range that’s a subset of $[0,1]$. Here, $f$ must have a range of $[0,1]$.

Suppose we define a function $f:[0,1]\to[0,1]$, such that using the Outer Lebesgue measure, the function is measurable in Caratheodory sense, and the range of the function is $[0,1]$.

Main Question

Can the pre-image of a non-measurable subset of $[0,1]$ under $f$, where the range of $f$ is $[0,1]$, be measurable?

Attempt:

I'm not sure how to approach this question due to my lack of formal training beyond Intro to Advanced Math. I assume if the points in the graph of $f$ is "spread out enough" in $[0,1]\times[0,1]$, there is some way to define $f$ where the pre-image of a non-measurable subset of $[0,1]$ under $f$ is measurable.

If this is not true, then the main question of this post doesn't give a function that satisfies the motivation of that post.

In case you want to read the motivation here, read the following:

I want to find a function $f:[0,1]\to[0,1]$ whose graph is dense, and somewhat but not too evenly distributed (i.e. with complete spaical randomness), in $[0,1]\times[0,1]$.

If the main question is wrong is there some way to fix the main question of the post to satisfy my motivation.

Arbuja
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1 Answers1

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You can take the Cantor function $f$: as shown in this post, for any set $A \subset [0,1]$, $f^{-1}(A)$ is Lebesgue-measurable.

The argument goes like this: $f^{-1}(A)$ is a union of at most countably many intervals (outside the Cantor set, where $f$ is locally constant), and a subset of the Cantor set (which has measure zero, and hence is measurable).

That's true in the particular case where $A$ is not measurable, which gives you an example.

charmd
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