# A converse to the First Fundamental Theorem of Calculus

I'm working on a proof of the following theorem:

"Let $f\colon [a,b]\to\mathbb{R}$ be monotone increasing, $F\colon [a,b]\to\mathbb{R}$ such that $F(x):=\int_{[a,x]} f$ and $x_0 \in [a,b]$. Then $F$ is differentiable at $x_0$ if and only if $f$ is continuous at $x_0$."

Now, the leftward implication follows easily from the First Fundamental Theorem of Calculus, but I'm having difficulties proving the rightward one; I've tried proof by contradiction but without much success for now. So, I'd appreciate any hint about how to prove this remaining implication.

Best regards,

lorenzo.

Suppose that $f$ is not continuous at $x_0$. Then, since $f$ is monotone increasing, $\lim_{x\to{x_0}^-}f(x)<f(x_0)$ or $\lim_{x\to{x_0}^+}f(x)>f(x_0)$. Let us suppose that the former assertion is correct and define $k=\lim_{x\to{x_0}^-}f(x)$. Then, if $x<x_0$,$$\frac{F(x)-F(x_0)}{x-x_0}=\frac{\int_x^{x_0}f(t)\,dt}{x_0-x}\leqslant\frac{k(x_0-x)}{x_0-x}=k.$$On the other hand, if $x>x_0$ then$$\frac{F(x)-F(x_0)}{x-x_0}=\frac{\int_{x_0}^xf(t)\,dt}{x-x_0}\geqslant\frac{f(x_0)(x-x_0)}{x-x_0}=f(x_0).$$So, the limit of $\frac{F(x)-F(x_0)}{x-x_0}$ at $x_0$ does not exist and therefore $F$ is not differentiable at $x_0$.
The case in which $\lim_{x\to{x_0}^+}f(x)>f(x_0)$ is similar.
• Using the fact that $f$ is continuous. – José Carlos Santos May 21 '17 at 18:50
• By hypothesis, $f$ is NOT continuous. – Mark Viola May 21 '17 at 18:55
• @lorenzo The fact that $f$ is not continuous at $x_0$ means that $\lim_{x\to x_0}f(x)$ either does not exists or that it exists and it is not $f(x_0)$. Since $f$ is monotone, the limit must exist. So, one of the limits $\lim_{x\to{x_0}^-}f(x)$ or $\lim_{x\to{x_0}^+}f(x)$ is not $f(x_0)$. Since $f$ increases, that means that $\lim_{x\to{x_0}^-}f(x)<f(x_0)$ or that $\lim_{x\to{x_0}^+}f(x)>f(x_0)$. – José Carlos Santos May 21 '17 at 21:57