just looking for a hint on how to proceed here.
In Tao's Analysis I, exercise 11.9.3 is as follows:
Let $a<b$ be real numbers, and let $f:[a,b]\rightarrow\mathbb{R}$ be a monotone increasing function. Let $F:[a,b]\rightarrow\mathbb{R}$ be the function $F(x):=\int_{[a,x]}f$. Let $x_0\in(a,b)$ be given. Show that $F$ is differentiable at $x_0$ iff $f$ is continuous at $x_0$. (Hint: one direction is taken care of by one of the Fundamental Theorems of Calculus. For the other, consider the left and right limits of $f$ and argue by contradiction.)
From this section, we are also given the two Fundamental Theorems of Calculus as follows:
(First Fundamental Theorem). Let $a<b$ be real numbers, and let $f:[a,b]\rightarrow\mathbb{R}$ be a Riemann integrable function. Let $F:[a,b]\rightarrow\mathbb{R}$ be the function $F(x):=\int_{[a,x]}f$. Then $F$ is continuous. Furthermore, if $x_0\in[a,b]$ and $f$ is continuous, then $F$ is differentiable at $x_0$, and $F'(x_0)=f(x_0)$.
(Second Fundamental Theorem). Let $a<b$ be real numbers, and let $f:[a,b]\rightarrow\mathbb{R}$ be a Riemann integrable function. If $F:[a,b]\rightarrow\mathbb{R}$ is an antiderivative of $f$, then $\int_{[a,b]}f=F(b)-F(a)$.
It was proven in an earlier section that if $f$ is monotone on a closed interval, it is integrable on that interval. Following the hint given, the First Fundamental Theorem tells us if $f$ is continuous at $x_0$, then $F(x):=\int_{[a,x]}f$ is differentiable at $x_0$, so we are done in that respect. My trouble here is with the other direction.
Following the hint again, suppose for the sake of contradiction that $F$ is differentiable at $x_0$, but $f$ is not continuous at $x_0$.
Since $f$ is a function on $[a,b]$, it is defined at $x_0$. Since $f$ is monotone increasing, $f(x_0)\geq f(x)$ for all $x<x_0$ and $f(x_0)\leq f(x)$ for all $x>x_0$. Therefore, $\lim\limits_{x\rightarrow x_0^-}f(x)\leq f(x_0)\leq\lim\limits_{x\rightarrow x_0^+}f(x)$, and since $f$ is not continuous, $\lim\limits_{x\rightarrow x_0^-}f(x)<\lim\limits_{x\rightarrow x_0^+}f(x)$.
This is where I seem to be stuck. My intuition is that now I should be able to show that $$\lim\limits_{x\rightarrow x_0^-}\frac{F(x)-F(x_0)}{x-x_0}<\lim\limits_{x\rightarrow x_0^+} \frac{F(x)-F(x_0)}{x-x_0},$$
but I can't quite piece together the connection. I've simplified the expression for which we're taking a limit to $$\frac{F(x)-F(x_0)}{x-x_0}=\frac{\int_{[x,x_0]}f}{x-x_0},$$
but this hasn't made anything clear for me, either.
Any hints on how to proceed would be much appreciated. Thanks!
Best Answer
The intuition is that if $f$ jumps at $x_0$, then the average of $f$ on intervals $(x_0 - h, x_0)$ to the left of $x_0$ will be smaller than the average on intervals $(x_0, x_0 + h)$ to the right of $x_0$.
Let $R = \inf_{x > x_0}f(x)$ be the right hand limit. Then $x > x_0 \implies f(x) \geq R$. Thus for $h > 0$, $\frac{1}{h}\int_{x_0}^{x_0 + h}f(t)\,dt \geq \frac{1}{h}Rh = R$. Get a similar estimate for $\frac{1}{h}\int_{x_0 - h}^{x_0}f(t)\,dt$ in terms of the left hand limit. Then let $h \to 0$.