# Problem related to Fundamental Theorem of Integral Calculus [duplicate]

This is a problem from the book of Terence Tao, Analysis I and it looks like-

Let $a<b$ be two real numbers, and let $f:[a,b]\to\Bbb{R}$ be a monotone increasing function. Let $F:[a,b]\to\Bbb{R}$ be the function $F(x)=\int_{[a,x]} f$. Let $x_0\in[a,b]$. Show that $F$ is differentiable at $x_0$ if and only if $f$ is continuous at $x_0$.

Observe that, the "if" part can be easily proved by using Fundamental Theorem of Calculus.
Just to recall ...
Statement of Fundamental Theorem of Calculus:

Let $a<b$ be two real numbers, and let $f:[a,b]\to\Bbb{R}$ be a Riemann Integrable function. Let $F:[a,b]\to\Bbb{R}$ be the function $F(x)=\int_{[a,x]} f$. Then $F$ is continuous on $[a,b]$. Furthermore, if $x_0\in[a,b]$ and $f$ is continuous at $x_0$, then $F$ is differentiable at $x_0$, and $F'(x_0)=f(x_0)$.

Here, $f$ is monotone increasing on $[a,b]$, hence $f$ is Riemann Integrable on $[a,b]$ and $f$ is continuous at $x_0\implies F$ is differentiable at $x_0$
I can't understand how to tackle the "only if" part. There is a hint given in the book: Consider left and right limits of $f$ and argue by contradiction.

Can anybody solve the part in a rigorous and analytical way?

## marked as duplicate by Paramanand Singh calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 17 '18 at 3:53

Suppose that $f$ is discontinuous at $x_0$. The limits $\lim_{x\to{x_0}^-}f(x)$ and $\lim_{x\to{x_0}^+}f(x)$ exist (since $f$ monotonic) and therefore, since $f$ is discontinuous at $x_0$ and $f$ is increasing,$$\lim_{x\to{x_0}^-}f(x)<\lim_{x\to{x_0}^+}f(x).$$If $h>0$,$$F(x_0+h)-F(x_0)=\int_{x_0}^{x_0+h}f(x)\,\mathrm dx\geqslant h\lim_{x\to{x_0}^-}f(x),$$and, if $h<0$,$$F(x_0+h)-F(x)=\int_{x_0}^{x_0+h}f(x)\,\mathrm dx\leqslant h\lim_{x\to{x_0}^-}f(x).$$So$$\lim_{h\to0^+}\frac{F(x_0+h)-F(x_0)}h>\lim_{h\to0^-}\frac{F(x_0+h)-F(x_0)}h$$and therefore $F$ is not differentiable at $x_0$.
It is essential that $f$ is monotonically increasing. Without this assumption you will find examples of functions $f$ which are not continuous at $x_0$, but $F$ being differentiable at $x_0$.
For a monotonically increasing $f$ it is well-known that, for any $x_0$, $f$ has a left limit $y_0^-$ and a right limit $y_0^+$. Clearly $f$ is continuous a $x_0$ if and only if $y_0^- = y_0^+$. Assume $f$ is not continuous at $x_0$. Then $y_0^- < y_0^+$. In this case $F(x_0+h) \ge F(x_0) + hy_0^+$ so that $F'(x_0) \ge y_0^+$, similarly $F(x_0) \le F(x_0-h) + hy_0^-$ so that $F'(x_0) \le y_0^-$ which is a contradiction.
• One can alternatively assume that $F'$ is differentiable at $x_0$. Then we obtain as above $y_0^+ \le F'(x_0) \le y_0^-$. This implies $y_0^- = y_0^+$ since we always have $y_0^- \le y_0^+$. – Paul Frost May 17 '18 at 8:07