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Could you provide an example of a function $f$ from $[0,1]^2$ to $[0,1]$ that is not Borel measurable? Is requiring $f$ to be Borel meusurable very restrictive?

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    $\begingroup$ Can you find a subset $A\subset[0,1]^2$ that is not Borel-measurable? Then the characteristic function of $A$ is not Borel-measurable. No, it is not very restrictive. Your question on itself testifies. $\endgroup$ – drhab May 3 '17 at 13:23
  • $\begingroup$ Thanks. To help me understanding better, could you give an example of $A$? $\endgroup$ – STF May 3 '17 at 13:26
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    $\begingroup$ Any Lebesgue-non measurabel set is not Borel-measurable (contraposition, Borel measurability implies the Lebesgue one). It could be for example, the Vitali set. Consider the relation $a\sim b\iff b-a$ is rational. Vitali set is the set of representants of cosets. $\endgroup$ – szw1710 May 3 '17 at 13:28
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    $\begingroup$ You could also take a look here $\endgroup$ – drhab May 3 '17 at 13:34

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