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Given $f:\mathbb{R}\rightarrow\mathbb{R}$ mapping each non-empty open set to the whole real line, is there always a set $A$ of measure zero such that the function $g:\mathbb{R}\rightarrow\mathbb{R}$ defined as $g(x)=f(x)$ on $A$ and $0$ everywhere else also maps every non-empty open set to the whole real line?

Edit: My idea is to use the sigmoid function to convert it into the equivalent problem about functions with domain $(0,1)$ and work in binary. Consider the set $B \subseteq (0,1)$ defined as the union over all $n$ of values in $(0,1)$ whose $n$th to $n+k$th binary places are 0, for some $k$ depending on $n$. This strategy lets us restrict to a set of arbirtrarily small positive measure. Can this idea be extended by incorporating the actual $f$ to get a let us restrict to a set of measure $0$?

Adam
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I think the following is a counterexample. Enumerate the intervals with rational endpoints as $I_1,I_2,I_3,\dots$. Constract pairwise disjoint nowhere dense closed sets $A_n\subset I_n$ with Lebesgue measure $|A_n|=a_n\gt0$. Let $h_n(x)=|A_n\cap(-\infty,x)|$ and let $B_n=\{x\in A_n:0\lt h_n(x)\lt a_n\}$. Finally, define $$f(x)=\tan\left(\frac{\pi h_n(x)}{a_n}-\frac\pi2\right)$$ if $x\in B_n$ and $f(x)=0$ if $x\in\mathbb R\setminus\bigcup_nB_n$.

Every nonempty open set is mapped onto the whole line. A nonempty open set contains some rational interval $I_n$ which contains $A_n$ which contains $B_n$. Now $h_n$ maps $A_n$ onto the closed interval $[0,a_n]$ so it maps $B_n$ anto the open interval $(0,a_n)$, which the function $x\mapsto\frac{\pi x}{a_n}-\frac\pi2$ maps onto the open interval $(-\frac\pi2,\frac\pi2)$ which the tangent function maps onto $\mathbb R$.

Every measure zero set is mapped to a measure zero set. Since the Lebesgue measure is countably additive, it will suffice to show that a measure zero subset of $B_n$ is mapped to a measure zero set. First, the function $h_n$ maps measure zero sets to measure zero sets, because $|h_n(J)|\le|J|$ for every interval $J$. Next, the function $x\mapsto\frac{\pi x}{a_n}-\frac\pi2$ maps measure zero sets to measure zero sets, because the image of an interval $J$ is an interval of length $\frac\pi{a_n}|J|$. Finally, the tangent function maps measure zero sets to measure zero sets, because it has a bounded derivative on every finite interval.

user14111
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  • Is this function also a counterexample? – Adam Jun 11 '24 at 03:17
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    I have no idea. – user14111 Jun 11 '24 at 04:43
  • I figured out that the function I gave in my other comment was not a counterexample, because you can always find a subsequence $a_n$ of the sequence in the sum where the sum of the subsequence converges, so we can restrict to the set of values of $x$ for which the $a_n$th binary place is $0$ for all $n$, and this set has measure $0$. – Adam Jun 11 '24 at 05:54
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    I don't understand your last comment (I don't doubt that you're right, I'm just slow), but I came back to say that I don't think the function in the answer you linked to is a counterexample because, if we define $A$ to be the set of numbers of the form $\sum_{n=1}^\infty x_n/2^n$ where $x_n\in{0,1}$ for all $n$ and $x_{2n}=x_{2n+1}$ for all but finitely many $n$, then I believe that $A$ has measure zero and $f(A\cap U)=\mathbb R$ for every nonempty open set $U$. – user14111 Jun 11 '24 at 06:27
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    OK, I think I get it now. You take a convergent subseries $\sum_na_n$ of the harmonic series, say $a_n=1/n^2$, and in that case you require the $k^\text{th}$ binary digit $x_k$ to be zero whenever $k$ is a square. Same idea as I had but neater. However, I think you want to allow a finite number of exceptions, your measure zero set will be dense; so $x_{n^2}=0$ for all sufficiently large $n$. – user14111 Jun 11 '24 at 08:05