Problem with left limits.

Question

Let $F: \mathbb{R} \to \mathbb{R}$ a non-decreasing function and suppose $G: \mathbb{R} \to \mathbb{R}$ defined by $G(x) = F(x+)$ (= the right limit of $F$ in $x$, which always exists for non-decreasing functions) is differentiable almost everywhere.

Is it true that $F$ is differentiable almost everywhere?

Here is the relevant fragment from the book that I'm reading:

Note that $F$ and $G$ are continuous at the same points and they agree at each point at which they are continuous; furthermore, if $F(x_0) = G(x_0)$, then $\frac{F(x)−F(x_0)}{x−x_0}$ lies between $\frac{G(x)−G(x_0)}{x−x0}$ and $\frac{G(x−)−G(x_0)}{x−x_0}$ . Hence if $G$ is differentiable at $x_0$, then $F$ is differentiable at $x_0$, and $F'(x_0) = G'(x_0)$. The almost everywhere differentiability of F follows.

I suspect the author uses the squeeze theorem here, but I see no reason why we should have that

$$\frac{G(x-)-G(x_0)}{x-x_0} \to G'(x_0)$$

Answer 1

Showing that $\frac {G(x-)-G(x_0)} {x-x_0} \to G'(x_0)$ as $ x \to x_0$: let $\epsilon >0$ and choose $\delta>0$ such that $|\frac {G(x)-G(x_0)} {x-x_0}- G'(x_0)|<\epsilon$ for $ |x-x_0| <\delta$. Then $|\frac {G(x-)-G(x_0)} {x-x_0}- G'(x_0)|\leq \epsilon$ for $ |x-x_0| <\delta$ by just taking left hand limits.

Answer 2

From scratch: Let $\epsilon>0$. There is a $\delta >0$ such that

$|x-x_0|<\delta\Rightarrow \left |\frac{G(x)-G(x_0)}{x-x_0}-G'(x_0)\right |<\epsilon.$

Fix an arbitrary $x\in (x_0-\delta,x_0+\delta).$ We will show that $\left |\frac{G(x-)-G(x_0)}{x-x_0}-G'(x_0)\right |\le\epsilon.$

So, choose $x>z\in (x_0-\delta,x_0+\delta)$. Then, $\left|\frac{G(z)-G(x_0)}{x-x_0}-G'(x_0)\right |<\epsilon.$ This is true for $all$ such $z$, and furthermore, we know that $\lim_{z\to x-}G(z)$ exists and of course, is what we are calling $G(x-).$

Thus, $\lim_{z\to x-}\left|\frac{G(z)-G(x_0)}{x-x_0}-G'(x_0)\right |=\left|\frac{G(x-)-G(x_0)}{x-x_0}-G'(x_0)\right |\le\epsilon.$