TITLE: How continuous can a bijection between line and plane be?
QUESTION [13 upvotes]: 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?

REPLY [17 votes]: 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 (not-continuous) 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.