In Folland's Real Analysis, theorem 3.23, the notation $F(x+)$ comes without definition as follows:
3.23 Theorem. Let $F: \mathbb{R} \to \mathbb{R} $ be increasing, and let $G(x) = F(x+)$.
The set of points at which $F$ is discontinuous is countable.
$F$ and $G$ are differentiable a.e., and $F'=G'$ a.e.
Proof. Since $F$ is increasing, the intervals $(F(x-), F(x+)) (x\in\mathbb{R})$ are disjoint, ...
So I hope to ask:
- What does $G(x) = F(x+)$ mean?
- One previous post says it means one-sided limit. If so, then why are intervals $(F(x-), F(x+)) (x\in\mathbb{R})$ disjoint?
My attempt 1:
Assume $x_1<x_2$, i.e., $x_2 = x_1 + \delta$ for some $\delta> 0$. Since $F$ is increasing, $F(x_2) = F(x_1) + \epsilon$ for some $\epsilon > 0$.$$F(x_1+) = \lim_{y\to x_1+}F(y) = \lim_{h>0, h\to 0}F(x_1+h) = F(x_1)+\varepsilon_1$$ $$F(x_2-) = \lim_{y\to x_2-}F(y) = \lim_{h>0, h\to 0}F(x_2-h) = F(x_2)-\varepsilon_2$$ where $\varepsilon_1>0, \varepsilon_2>0$.
Question: I think $\varepsilon_1$ and $\varepsilon_2$ should be $0$, but that seems to require the continuity of $F$, which is not assumed. Am I going in the right direction?
My attempt 2:
Assume $x_1<x_2$, and there exists $y\in\mathbb{R}$ s.t. $x_1<y<x_2$ since $\mathbb{R}$ is dense. We have $$F(x+) = \lim_{z\searrow x}F(z) = \inf_{z>x}F(z)$$ $$F(x-) = \lim_{z\nearrow x}F(z) = \sup_{z<x}F(z)$$ Thus for any $x_1<y<x_2$: $$F(x_2-) = \sup_{z<x_2}F(z) \geq F(y)$$ $$F(x_1+) = \inf_{z>x_1}F(z) \leq F(y)$$ $$F(x_2-) - F(x_1+) \geq F(y) - F(y) = 0$$ So we are done.
Question: Is this attempt correct?
Any help would be appreciated!
