I want to find some subset of R^2 which its intersection with every vertical line is measure zero if we see it as a subset of R and it is not measure zero in R^2?
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2$\begingroup$ Why?${}{}{}{}{}$ $\endgroup$– Asaf Karagila ♦Commented Mar 14, 2014 at 11:27
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$\begingroup$ because of some reasons in integrability of multivariable functions $\endgroup$– alichCommented Mar 14, 2014 at 11:29
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7$\begingroup$ Wouldn't this contradict Fubini's theorem? Of course, assuming that the set is measurable in the first place. $\endgroup$– Alex DegtyarevCommented Mar 14, 2014 at 11:31
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$\begingroup$ Why does It contradict Fubini theorem? $\endgroup$– alichCommented Mar 14, 2014 at 11:35
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$\begingroup$ @alich: Because such a set $A\subseteq\mathbb{R}^2$ would satisfy $0\neq \lambda^2(A) = \int_{\mathbb{R}^2} \chi_A(x,y) d\lambda^2(x,y) = \int_\mathbb{R} \int_\mathbb{R} \chi_A(x,y) d\lambda^1(x) d\lambda^1(y) = \int_\mathbb{R} 0 d\lambda^1(y) = 0$. (where $\lambda^{1,2}$ denotes the 1-dimensional and 2-dimensional lebesgue meausure respectively) $\endgroup$– Johannes HahnCommented Mar 14, 2014 at 11:59
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1 Answer
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It is a well known result of Sierpinski that there exists non-measurable subsets of $\mathbb{R}^{2}$ which intersect each line in at most two points. Furthermore, there exists a real-valued function whose graph is a non-measurable subset of $\mathbb{R}^{2}$.
W. Sierpi´nski: Sur un probl`eme concernant les ensembles mesurables superficiellement. Fund. Math. 1 (1920), 112–115.
Gelbaum, Bernard R., and John M. H. Olmsted. Counterexamples in Analysis. San Francisco: Holden-Day, 1964. (p. 142-145)
