A bounded variation (BV) function $f$ on the interval $[a,b]$ can be written as a difference of two monotone increasing function. This can be done by a construction where $$F(x):=\sup \sum_{j=1}^{n-1}|f(x_{j+1})-f(x_j)|$$ where the supremum is taken over the $x_1,\ldots,x_n$ which satisfy $a=x_1<x_2<\ldots<x_n=x$, and then writing $f=F+(f-F)$.
What about a BV function on the whole $\mathbb{R}$? Can it also be written as a difference of two monotone increasing function?
Fishing out another answer from a comment: yes, $f=F+(f-F)$ works on $\mathbb R$ with $$F(x) = \begin{cases}V(f,[0,x]) &, x \geqslant 0\\ -V(f,[x,0]) &, x < 0. \end{cases}$$
A slightly different approach (which leads to the same thing) is to begin by writing $$f(x)=f(0)+(f(x)-f(0))\chi_{(0,\infty)}+(f(x)-f(0))\chi_{(-\infty,0)}$$ The first term is constant. Each of the other two is $0$ on half-axis. On the other half-axis, the construction in your post applies verbatim.
