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?
