Is there a bijection $f$ from $[0, 1]$ to $[0, 1]^2$ such that the set of points of discontinuity of $f$ has measure zero? If not, could it be dense/comeager?

3$\begingroup$ Somewhat relevant (to my mind) is a paper, On Jordan arcs and Lipschitz classes of functions defined on them (projecteuclid.org/download/pdf_1/euclid.acta/1485889215 ), by Besicovitch and Schoenberg from 1961 in Acta Math., in which they probe the extent to which $[0,1]^2$ may be 'approximated' by an injective continuous image of $[0,1]$. It includes both positive (Theorem 1) and negative (Theorem 2) results. $\endgroup$– Vesselin DimitrovJul 1 '18 at 21:37
Yes! Let $X$ be the Cantor middle third set. $X \subset [0,1]$ is closed, measure zero, and $X = 2^{\aleph_0} = X^c$.
Notice there is a continuous bijection $g: X^c \rightarrow (X^c \times \{0\})$ given by $g(x) = (x,0)$. It's easy to see the caridnality of $(X^c \times \{0\})^c$ is $2^{\aleph_0}$, and $X$ has cardinality $2^{\aleph_0}$, so there is also a (notcontinuous) bijection $h: X \rightarrow (X^c \times \{0\})^c$.
Now define $f(x) = g(x)$ if $x \not\in X$ and $f(x) = h(x)$ if $x \in X$. We can check that $f$ is continuous on $X^c$ which is an open measure 1 set.
It might be more interesting to know about the set of points of discontinuity of $f^{1}$. Perhaps a property similar to Lebesgue covering dimension can give some constraints.

$\begingroup$ I'm having trouble seeing how this is a bijection between $[0,1]$ and $[0,1]^2$. Where does $X\times X$ come in? $\endgroup$– MitchJul 2 '18 at 12:58

$\begingroup$ $[0,1] = X \cup X^c$, and $[0,1]^2 = (X^c \times \{0\}) \cup (X^c \times \{0\})^c$ are partitions of $[0,1]$ and $[0,1]^2$ respectively (here $(X^c \times \{0\})^c$ is the complement in $[0,1]^2$). $h$ and $g$ are bijections between corresonding peices, so $f$ is also a bijection. $\endgroup$– JamesJul 2 '18 at 13:24

$\begingroup$ Oh OK. But I don't see how $(X^c\times\{0\})\cup (X^c\times\{0\})^c$ covers all of $[0,1]^2$. $\endgroup$– MitchJul 2 '18 at 13:44

$\begingroup$ @Mitch The c means complement, so it covers the square, by definition, because the second piece is the complement of the first piece. The point of James's nice answer is that there is a trivial continuous injection on the complement of the Cantor set, and then one uses the rest, the Cantor set itself, to extend it to a bijection. This is possible because the Cantor set, although measure zero and meager, has size continuum. This makes for a bijection of $[0,1]$ with $[0,1]^2$ that is continuous on a measure 1 open set. $\endgroup$ Jul 2 '18 at 15:39

1$\begingroup$ I took $(X^c\times\{0\})^c$ to mean the complement in the square, so of course this covers the rest of the square. $\endgroup$ Jul 2 '18 at 15:53