Borel Sets and Borel Functions

Question

I read the following argument in Billingsley (1995). Consider a function $F(x)$ and define four associated quantities at $x$:

$$ D^F(x) := \limsup_{h\downarrow 0} \frac{F(x+h)-F(x)}{h}; \\ D_F(x) := \liminf_{h\downarrow 0} \frac{F(x+h)-F(x)}{h}; \\ ^{F}D(x) := \limsup_{h\downarrow 0} \frac{F(x)-F(x-h)}{h}; \\ _{F}D(x) := \liminf_{h\downarrow 0} \frac{F(x)-F(x-h)}{h}. $$

It is clear that $F$ has a derivative at $x$ if and only if the above four quantities exist and equal. Now assume that $F$ is measurable and non-decreasing and hence has at most countably many discontinuity points. Let $M$ be a countable and dense set containing all these discontinuity points. Moreover, let $r_n(x)$ be the smallest number of the form $\frac{k}{n}$ exceeding $x$. Then it claims that:

$$ D^F(x) = \lim_{n\to\infty} \sup_{\substack{x<y<r_n(x) \\ y\in M}} \frac{F(y)-F(x)}{y-x}. $$

Furthermore, it is claimed that the function inside the limit is measurable because the $x$-set where it exceeds $\alpha$ is

$$ \cup_{y\in M} \{x: x<y<r_n(x), F(y)-F(x)>\alpha(y-x)\}. (*) $$

Therefore, it concludes that $D^F(x)$ is measurable but it may well be infinite. Similarly the other three quantities $D_F(x), ^{F}D(x)$ and $_{F}D(x)$ are measurable, too. Finally, the set where the four quantities have a common finite value is a Borel set.

My questions are as follows:

1. I can see why $(*)$ are true. But it took me a while to figure it out. That is, I cannot think of an intuitive way to explain it and remember it. Does anyone have a clever way to make sense of it, please?
2. I still could not see why $D^F(x)$ is measurable. Could anyone explain it, please?
3. Finally, why is the set where the four quantities have a common finite value is Borel, please? Thank you!

Answer 1

(1) Since a limit exists iff $\limsup=\liminf$ and the derivative exists iff left derivative = right derivative. Thus you have $2\times 2=4$ limits.

(2) $D^F(x)=\lim_{n→∞}\sup _{x<t<r_n(x)}\frac{F(t)−F(x)}{t−x} = \lim_{n→∞}\sup_{x<y<r_n(x),y∈M}\frac{F(y)−F(x)}{y−x}.$ The first equality can be considered as definition or an equivalent way of formulating the $\limsup$. The second equality is because $M$ is dense and contains all the discontinuous points and $F$ is continuous a.e. (try to take an sequence $t_m$ approaching $\sup _{x<t<r_n(x)}\frac{F(t)−F(x)}{t−x}$ ($n$ ,$x$ fixed) , and you will see why the second equality is true.).

Since $\lim_{n\rightarrow \infty}$ operation still gives us (Borel)measurable function, we only need to show $\sup_{x<y<r_n(x),y∈M}\frac{F(y)−F(x)}{y−x} $ is measurable, which is the same as showing $\cup_{y\in M}\{x: x<y<r_n(x), F(y)-F(x)>\alpha(y-x)\}$ is measurable. As $\frac{F(y)-F(x)}{y-x}$ is measurable for fixed $y$ and there is countably many $y$, we see $(*)$ is a measurable set and thus $D^F$ is measurable.

(3) The set for four quantities have a common finite value is Borel since you could write the set as $\{x: D_F(x)=D^F(x)\}\cap \{x: D_F(x)=^FD(x)\}\cap\{x:D_F(x)=_FD(x)\}\cap\{x: D_F(x)<\infty\}\cap\{x: D^F(x)<\infty\}\cap\{x: _FD(x)<\infty\}\cap\{x: ^FD(x)<\infty\}$. Since $D$s are all Borel measurable, the sets are all Borel measurable and so is their intersection.

Are you self-learning measure-theoretic probability ?