When is a sum of jump functions still a jump function?

Question

Setup

Let's adopt the definition at https://encyclopediaofmath.org/wiki/Jump_function:

• A right-continuous function of bounded variation $f:[0,1]\to\mathbb R$ is a jump function if $$f(x)=\sum_{y\leq x}f(y^+)-f(y^-).$$

These functions occur as the discontinuous "part" of a general BV function in the Lebesgue decomposition. They are a special case of càdlàg functions, being right-continuous and having left limits everywhere. They are more general than step functions, whose jumps have to occur on a finite set (or, depending on the author, at least a discrete set).

For example, the below image depicts the function $f$ that has, for all $n\geq1$, a jump at $\{\varphi n\}$ of size $(-1)^n/n^2$, where $\varphi=(\sqrt5-1)/2$ is irrational, so there are infinitely many jumps between $f(0)=0$ and $f(1)=-\eta(2)=-\pi^2/12$. The total variation of $f$ is $TV(f)=\zeta(2)=\pi^2/6$ per Euler. The plotting software makes discontinuities look like vertical lines, but you get the idea. This example should make it clear that a jump function is not necessarily constant, or monotone, or continuous, over any open interval.

Question

Given that notion, two sub-questions:

1. Given an infinite series of jump functions $a_n$, when is it true that $S=\sum_{n=1}^\infty a_n$ is a jump function?
2. When is it also true that the set of discontinuities of $S$ is contained in the union of the sets of discontinuities of the $a_n$ (without taking any closures)?

Known results

I'm aware of the following arguments:

• If $\sum a_n$ converges absolutely in the BV norm $\|f\|_{BV}=\|f\|_{L^1}+TV(f)$, then we get both (1) and (2). This is because the total variation $TV(a_n)$ is just the sum of absolute values of the jumps, and then $S$ is a double sum; by Fubini, we can exchange the order of summation.
• If $\sum a_n$ merely converges uniformly, then we get nothing. $S$ might be an absolutely continuous function like $S(x)=x$. It might be a Cantor staircase. There are lots of possibilities. I guess that's convenient if you're inventing measure theory, but for this question, we're trying to avoid such mutations.

Is there a nice condition, somewhere in between the above, that is necessary and sufficient?

Bonus points if there's literature to cite.

(This question is motivated by an application of mine, where I know the locations of the discontinuities of $a_n$, and I want to write down $S$. But I'm lazy, and I don't want to bother to measure the sizes of the jumps of each $a_n$, and then add them all up and do a bunch of index chasing. Instead, I just want to write down $S$, which happens to obey a nice functional equation, such that I can very easily measure the size of each jump, provided that I know where to look. I do have absolute convergence in BV, so my application will work. I'm just asking this question out of curiosity.)

Answer 1

I give two examples where the properties in questions 1 and 2 do not hold.

1. Define $f_n(x)=2^{-n}\lfloor 2^nx\rfloor$ and $a_n=f_{n+1}-f_n$ on the interval $[0,1]$.
   Both $f_n$ and $a_n$ are step functions, but $\sum_n a_n=\lim_n f_n$ is the identity function on $[0,1]$, which is not a jump function.
   The convergence is uniform but not in the BV norm since the total variation of $a_n$ is constant of value $1$.
2. For the second question, take $a_n$ the characteristic function of the interval $\displaystyle\left[\frac 12-\frac{1}{2^n},\frac 12-\frac{1}{2^{n+1}}\right)$ for all $n>0$.
   Then $\sum_n a_n$ is the characteristic function of $[0,\frac 12)$, whose discontinuity is not contained in the union of discontinuities of $a_n$.

Now the second example is in fact a counter-example to the backwards implication:
convergence in the BV-norm $\Leftarrow$ convergence to a jump function