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 interseting to know about the set of points of discotninuity of $f^{-1}$. Perhaps a property similar to Lebegue covering dimension can give some constraints.

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.

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 $2^{\omega}$, and $X$ has cardinality $2^{\omega}$, 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.

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 interseting to know about the set of points of discotninuity of $f^{-1}$. Perhaps a property similar to Lebegue covering dimension can give some constraints.