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We know that there does not exist a continuous bijective function from $[0,1]$ to $[0,1]^2$. (More generally, $[0,1]$ is not homeomorphic to $[0,1]^2$.) However, is there a continuous a.e. (almost everywhere) bijective function from $[0,1]$ to $[0,1]^2$?

The motivation for asking this question comes from the following:

1) $([0,1],B_{[0,1]})$ is borel isomorphic to $([0,1]^2,B_{[0,1]^2})$, i.e., there exists a measurable bijective function $f$ from $[0,1]$ to $[0,1]^2$, and its inverse $f^{-1}$ is also measurable.

2) By Lusin's theorem, we have that for every $\varepsilon > 0$, there exists a compact $E ⊂ [0, 1]$ such that $f$ restricted to $E$ is continuous almost everywhere and $\mu (E)>1-\varepsilon $, where $\mu$ is the Lebesgue measure.

So the answer for my question seems to be positive, at least for the case with $[0,1]$ and $[0,1]^2$ replaced by some 1-D space $E$ and 2-D space $E'$. Is there anybody can prove it or disprove it?

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    $\begingroup$ You have stated wrongly the definition of homeomorphism. An homeomorphism is a bicontinuous bijective function. $\endgroup$ – Alessandro Contini Jun 26 '18 at 9:17
  • $\begingroup$ @AlessandroContini Thank you for pointing out the mistake! The question has been revised. $\endgroup$ – Lay Jun 26 '18 at 9:23
  • $\begingroup$ en.wikipedia.org/wiki/Space-filling_curve Maybe this will help a little, though it didn't anwser the question. $\endgroup$ – XIAODA QU Jun 26 '18 at 9:44
  • $\begingroup$ @XIAODAQU I am looking for a bijective function. The Space-filling curve is only injective. $\endgroup$ – Lay Jun 26 '18 at 11:29
  • $\begingroup$ @LeiYu Yeah I'm aware of that. And, do you have any materials about the borel isomorphic between $([0,1],\mathcal{B})$ and $([0,1]^2,\mathcal{B})$? I never heard of this before. $\endgroup$ – XIAODA QU Jun 26 '18 at 12:04

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