What is the completition of the step functions on an interval with respect to the BV norm? I am interested in the closure of the space of step functions
on a bounded interval with respect to the BV norm. I understood that
the closure is a proper subspace of BV but has this space a name? Is it the
space of special BV functions?
 A: It's the space of BV functions whose weak derivative is a discrete measure. Equivalently, the BV functions with zero derivative except on a countable set.
The space of right continuous functions in $BV([0,1])$ is bi-Lipschitz equivalent to the space of signed measures on $[0,1]$ equipped with the total variation norm. A measure $\mu$ corresponds to the function $f(x)=\mu([0,x]);$ the BV norm is at most twice the total variation of the measure. It is not hard to see that the completion of the finitely supported signed measures is the discrete signed measures.
For the classical pointwise definition of BV functions, not necessarily right-continuous, we need to add an absolutely summable combination of delta functions. This can be acheived using degenerate intervals.
In one dimension SBV is the space of BV functions whose weak derivative is the sum of an absolutely continuous and a discrete measure i.e. no singular continuous part in the Lebesgue decomposition. This description suggests that functions with discrete weak derivative are a proper subspace. For example by considering the total variation of the derivative it is easy to see that the element of $BV([0,1])$ defined by $f(x)=x$ will have BV norm at least $1$ from any step function.
