Is it possible for there to be an $X\subset\Bbb R^2$ such that for all lines $\ell$ in the plane, we have that $X\cap\ell$ has measure $1$ in $\ell$?
I don't know enough measure theory to see how to figure it out.
Subset of plane with measure $1$ in all lines
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$\begingroup$
measure-theory
set-theory
lebesgue-measure
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0Surely such a set must be quite complicated: What is can be said is that if it is Lebesgue measurable, then its measure is $\infty$ (by Cavalieri's principle). In particular it cannot be bounded. – 2017-01-09
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0It also cannot be bounded because it has nonempty intersection with all possible lines, even the ones that are really far away – 2017-01-09
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0Now posted to MO, http://mathoverflow.net/questions/261267/a-subset-of-plane-that-meets-every-line-at-length-one – 2017-02-03
1 Answers
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Under CH, there is such a set. To construct it, list all lines $\{L_i : i < \omega_1\}$ and construct $S$ inductively as follows. Start with empty $S$. At stage $i$, suppose $S$ is contained in $\bigcup \{L_j : j < i\}$ and it meets every $L_j$ at length one, for $j < i$. Note that $L_i \cap \bigcup \{L_j : j < i\}$ is countable. So we can choose a compact subset contained in $L_i$ of length one which is disjoint with $\bigcup \{L_j : j < i\}$ and add it to $S$. Of course this $S$ is not measurable but Theorem 7 here of Kolountzakis and Papadimitrakis shows it can't be.
Note that this construction also works as long as all sets of size less than continuum are null. But I don't immediately see how to do this in ZFC alone.