If the function $f:R\to R$ is of BV class then it has at most countably many discontinuity points (since it can be represented as a sum of two monotonic function).
I am interested to know whether the BV function $f:R^2\to R$ has at most countable many curves along which it is discontinuous?
Thanks a lot in advance.
The answer seems to be negative.
For a counterexample, let $T$ be a continuous infinite binary tree in the unit disk, where one finds the branching nodes as one moves away from the origin and toward the boundary of the unit disk. I imagine that one starts at the origin and finds the branching nodes of the tree as one moves way from the origin towards the edge of the unit circle. For example, perhaps the $k^{th}$ branching nodes all lie on the circle of radius $1-2^{-k}$, which then continue outwards to the next circle. This tree divides the unit disk into countably many regions, such that any disk at the origin of radius less than $1$ meets only finitely many of them. Let $f$ be a function that decays to $0$ at the unit circle (and outside it), and is discontinuous exactly at the points on this tree. For example, perhaps crossing the tree at the $k^{th}$ circle causes a jump of size about $2^{-k}$.
If I am not mistaken, this function has bounded variation (I would welcome confirmation of this by functional analysis experts). But meanwhile, since there are continuum many paths through the tree, there are continuum many (and hence uncountably many) curves, each of which consists entirely of points of discontinuity of $f$. Each such curve travels from the origin to the boundary of the unit disk, and any two of them deviate from each other at some radius strictly less than $1$.
