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For $f: [0,1] \to [0,1] $,

Define $A_f = \{y : |\{f^{-1}(y)\}| > \aleph_0\}$. It is known that $A_f$ must have null measure if $f$ is continuous and of bounded variation. This thread explains the theorem clearly. My question is whether every null-measure subset of $[0,1]$ is attained as $A_f$ for some $f$ of bounded variation, even better if it's continuous. An insightful example to understand would be $A_f = [0,1] \cap (\mathbb{Q} + C)$, the rational translates of a thin Cantor set in $[0,1]$.

Cases I can deal with:

If $A_f$ is countable, say $\mathbb{Q}$, the construction is easy. We can take some variant of Cantor's Staircase function, which is monotonic so trivially of bounded variation.

The case of $A_f$ an arbitrary compact set is more difficult. The details are long, but broadly my method involves taking a particular monotonic function with the desired range and a restricted closed domain $X$, then subtly adjusting it to get $f$ which has the desired $A_f$ but is still continuous and of bounded variation. $D$ is chosen so that there is a known separable subset $S$, and we compute the variation by examining the variation of $f$ on finite subsets of $S$. We can then extend $f$ continuously to all of $[0,1]$ by linearity. However, the construction of $D$ that I used is dependent on $A_f$ being closed, so I do not see this method generalising to $A_f$ an arbitrary null measure set in $[0,1]$.

shyb
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    A simple counting argument shows that not all Lebesgue measure zero subsets of $[0,1]$ can be an $A_f$ set for a BV function $f:[0,1] \rightarrow \mathbb R$ -- There are $2^{\mathfrak c}$ many Lebesgue measure subsets of $[0,1]$ and only $\mathfrak c$ many BV functions (even including discontinuous) and hence only $\mathfrak c$ many $A_f$ sets. Probably any $A_f$ set has a fairly low Borel structure for an arbitrary function (e.g. my guess is at most $F_{\sigma \delta}$ or $G_{\delta \sigma}),$ and it's also likely that this is sharp even for BV functions, but I don't know any references. – Dave L. Renfro Sep 16 '23 at 13:06
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    (follow-up to previous comment) By "this is sharp even for BV functions", I mean that each Lebesgue measure zero subset of $[0,1]$ satisfying a certain Borel condition (to be determined) is an $A_f$ set for some BV function $f:[0,1] \rightarrow \mathbb R.$ – Dave L. Renfro Sep 16 '23 at 13:21
  • Thank you, I hadn't thought about it from that point of view! I suppose a sensible next step would be to try proving it is possible for all null-measure $F_\sigma$ and $G_\delta$ subsets of $[0,1]$. – shyb Sep 16 '23 at 14:41

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