proof related to Fundamental Theorem of Calculus $f$ is not necessarily continuous on an interval I want to solve the problems from the book "A Course of Modern Analysis" (P 65) by E.T. Whittaker & G.N. Watson. The notation in this book is a little old-fashioned and I don't really like their notation.

If $f(x)$ is integrable when $a_1 \le x \le b_1$, when $a_1 \le a < b < b_1,$ we write
$$
\int_{a}^bf(x)dx = \phi(a,b),
$$
and if $f(b+0)$ exists, then
$$
\lim_{\delta \to 0^+} \frac{\phi(a, b+\delta)-\phi(a, b)}{\delta} = f(b+0).
$$

Here is my attempt for the problem:

\begin{align*}
\lim_{\delta \to 0^+} \frac{\phi(a, b+\delta)-\phi(a, b)}{\delta}  &= \lim_{\delta \to 0^+} \frac{\int_{a}^{b+ \delta}f(x)dx -\int_{a}^bf(x)dx }{\delta} \\ 
&=  \lim_{\delta \to 0^+} \frac{\int_{b}^{b + \delta}f(x)dx}{\delta}.
\end{align*}
Then I got stuck here.

 A: This is a general version of the Fundamental Theorem of Calculus.
Just note that $f(x) \to f(b+0)=L(\text{say}) $ as $x\to b^+$ and thus we have corresponding to a given $\epsilon>0$ a $\delta>0$ such that $$L-\epsilon <f(x) <L+\epsilon $$ whenever $b<x<b+\delta$. If $0<h<\delta$ then integrating the above inequality over interval $[b, b+h] $ we get $$h(L-\epsilon) <\int_b^{b+h} f(x)\, dx<h(L+\epsilon) $$ ie $$L-\epsilon<\frac{1}{h}\int_b^{b+h}f(x)\,dx<L+\epsilon $$ for all $h$ with $0<h<\delta$. This means (by definition of limit) that $$\lim_{h\to 0^+}\frac{1}{h}\int_b^{b+h}f(x)\,dx=L$$ which is exactly what you wanted to prove.
A: \begin{align*}
& \lim_{\delta \to 0^+} \frac{\phi(a, b+\delta)-\phi(a, b)}{\delta} \\
&= \lim_{\delta \to 0^+} \frac{\int_{a}^{b+ \delta}f(x)dx -\int_{a}^bf(x)dx }{\delta} \\ 
&= \lim_{\delta \to 0^+} \frac{\int_{b}^{b + \delta}f(x)dx}{\delta} \\
&= \lim_{\delta \to 0^+} \frac{\lim\limits_{n\to+\infty}\sum\limits_{i=1}^{n} f(b+i\delta/n) \, \delta/n}{\delta} \tag{$f$ is integrable on $[a,b_1]$} \\
&= \lim_{\delta \to 0^+} \lim_{n\to+\infty} \frac1n \sum_{i=1}^{n} f(b+i\delta/n) \\
&= f(b+) \tag{claim to be proved}
\end{align*}
To prove the claim, let $\epsilon > 0$.  Using the existence of $f(b+)$ to establish $\delta > 0$ such that $|f(x)-f(b+)| < \epsilon$ whenever $x \in (b,b+\delta)$.
\begin{align}
\left(\frac1n \sum_{i=1}^{n} f(b+i\delta/n)\right) - f(b+) &= \left[\frac1n\sum_{i=1}^{n-1} (f(b+i\delta/n)-f(b+))\right] + \frac{f(b+\delta)-f(b+)}{n} \\
\left \lvert \left(\frac1n \sum_{i=1}^{n} f(b+i\delta/n)\right) - f(b+) \right \rvert &< \frac{n-1}{n} \,\epsilon + \left \lvert \frac{f(b+\delta)-f(b+)}{n} \right \rvert \\
&< \epsilon + \left \lvert \frac{f(b+\delta)-f(b+)}{n} \right \rvert
\end{align}
Take $N > \left\lceil\dfrac{f(b+\delta) - f(b+)}{\epsilon}\right\rceil$.  Whenever $n \ge N$, then
$$\left \lvert \left(\frac1n \sum_{i=1}^{n} f(b+i\delta/n)\right) - f(b+) \right \rvert < 2 \epsilon.$$
