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Functions like Conway's base-13 function map each open set to the whole real line. Functions like this one I think map every set of positive Lebesgue measure to the real line but I don't know how to prove it. Can this result be extended in the way described in the question? If it does not, then the only other functions with this property use Axiom of Choice.

Adam
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    No, there is none. Take an arbitrary function from $R$ to $R$, and consider the preimages of the intervals $]-\infty, 0]$ and $[0,\infty[$. At least one has the same cardinality as $R$. – Alexey Jun 08 '24 at 22:57
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    This has already been discussed in a different question of the OP’s, but for the sake of correcting the record, let me note here that the function in the second link does *not* map every set of positive measure onto the entire real line. Indeed, the function in the link is measurable, but any function that maps every set of positive measure onto the entire real line cannot be measurable. – David Gao Jun 09 '24 at 06:56
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    (Also, a reminder that you should accept an answer if you find it satisfactory.) – David Gao Jun 09 '24 at 06:59
  • This is a duplicate. – MathematicsStudent1122 Sep 05 '24 at 13:21

1 Answers1

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Suppose such a function $f$ existed. Since $f$ is surjective, the pre-image of $\mathbb{R}\setminus\{0\},$ $X$, has cardinality $\vert \mathbb{R}\vert,$ and so we must have $f(X)=\mathbb{R}.$ But $f(X) = \mathbb{R}\setminus\{0\}.$ A contradiction has arisen.

Adam Rubinson
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