Decomposition of function of bounded variation

Question

Suppose we have $f:\mathbb{R} \rightarrow \mathbb{R}$ which is of bounded variation. I would like to show that it can be presented as a sum of left and right continuous functions of bounded variation. Actually I have no idea how to prove that since I think it can't be proven that Jordan decomposition has that properties and direct constructions failed in a way I can't prove them to be right.

Answer 1

My guess is that the OP has already solved the problem. For other readers irritated by cryptic hints, here are some more details.

As already indicated the problem is really just a refinement of the usual version one finds:

A. If $f:[a,b]\to \mathbb R$ has bounded variation then $$f(x) = g(x) + s(x)$$ where $g$ is a continuous function of bounded variation and $s$ is a saltus function.

Instead, we ask as well for the saltus part to be split into two nicer pieces:

B. If $f:[a,b]\to \mathbb R$ has bounded variation then $$f(x) = g(x) + s_L(x) + s_R(x)$$ where $g$ is a continuous function of bounded variation and $s_R$ is a right-continuous saltus function and $s_L$ is a left-continuous saltus function.

Let $\{x_n\}$ be the sequence of discontinuities of $f$ and let $a_n=f(x_n)-f(x_n-)$ and $b_n=f(x_n+)-f(x_n)$ be the two one-sided jumps. (One or both of these are nonzero.)

Write $I(x)=0$ for $x<0$ and $I(x)=1$ for $x\geq 0$. Write $J(x)=0$ for $x\leq 0$ and $J(x)=1$ for $x> 0$.

Then the saltus function for $f$ is the function

$$s(x)= \sum_{n=1}^\infty \left\{ a_n I(x-x_n)+ b_nJ(x-x_n)\right\}.$$

The usual textbook proof one can find checks the details to see that $s$ is well-defined, that $f-s$ is continuous and has bounded variation. And $s$ is a saltus function by definition. This is probably the standard proof for A.

To get the version B just split the series into two series; one half is a right-continuous saltus $s_R$ and the other is a left-continuous saltus $s_L$.