Math people:
This question is related to how can you "fix" one of the definitions of a BV function of one variable? . Suppose $f \in BV([0,1])$. I really have two-three questions. The second only can be answered if the answer to the first is "yes". I will accept an answer to just the first if no one can answer all of them.
(1) Does $$\lim_{t \to 0^+} \frac{\int_x^{x+t} f(s)\,ds}{t}$$ exist for all $x \in [0, 1)$?
(2) If the answer to (1) is "yes", and I define $g:[0,1] \to \mathbb{R}$ by
$$ g(x) = \lim_{t \to 0^+} \frac{\int_x^{x+t} f(s)\,ds}{t} \hbox{ for } x < 1,\ g(1)= \lim_{t \to 0^+} \frac{\int_{1-t}^{1} f(s)\,ds}{t},$$
is $g=f$ Lebesgue-a.e.? And is $g$ right-continuous? A reference would be appreciated. I would guess that all are true and known, but I can't find a reference.
Stefan (STack Exchange FAN)
The answer to both questions is "yes", if you accept that $f$ is a.e. the sum of a nondecreasing function $h_1$ and a nonincreasing function $h_2$, as is well-known. Then, for all $x \in [0,1)$,
$$\lim_{t \to 0^+} \frac{\int_x^{x+t}f(s)\,ds}{t} = \lim_{t \to 0^+} \frac{\int_x^{x+t}h_1(s)\,ds}{t} + \lim_{t \to 0^+} \frac{\int_x^{x+t}h_2(s)\,ds}{t} = \lim_{x' \to x^+} h_1(x') + \lim_{x' \to x^+} h_2(x')$$
and the limits on the right exist because $h_1$ and $h_2$ are monotone. Since $h_1$ and $h_2$ are monotone, they have at most countably many discontinuities, and (2) follows.