The lemma will show that
If $f$ is bounded and measurable on $[a,b]$ and $F(x) = \int^x_a f(t)dt + F(a)$, then $F'(x)= f(x)$ for almost all $x$ in $[a,b]$. I understand the basic idea of the proof, however, did not figure out some details as below in the proof.
On page 107, the proof states that $\lim_{h\to0} \frac{1}{h} \int^c_a (F(x+h) - F(x))\,dx = \lim_{h \to 0} \left[ \frac{1}{h} \int^{c+h}_c F(x)\,dx - \frac{1}{h} \int^{a+h}_a F(x)\,dx \right]= F(c) - F(a)$, where $h = \frac{1}{n}$ and $n$ is a natural number.
I did not get the idea why we could get $\lim_{h \to 0} \frac{1}{h} \int^{c+h}_c F(x)\,dx = F(c)$ and $\lim_{h \to 0} \frac{1}{h} \int^{a+h}_a F(x)\,dx = F(a)$.
My guess is that it is due to the fact that $F(x)$ is continuous, however, I cannot fill the gap, and not sure my guess is correct. Any help? The complete proof is attached below.
Let's prove that $\lim_{h \to 0}\frac1h \int_c^{c+h}F(x)\,dx=F(c)$, using the fact that $F$ is continuous.
Without loss of generality we can assume that $h>0$ (the same argument works for $h$ negative). Since $F$ is continuous on $[c,c+h]$, there exists two points $a,b \in [c,c+h]$ such that $F(a)=m$ and $F(b)=M$, where $m$ and and $M$ are, respectively, the absolute minimum and maximum of $F$. Then we have $$ mh \leq \int_c^{c+h}F(x)\,dx \leq Mh $$ which is the same as $$ F(a) \leq \frac1h \int_c^{c+h}F(x)\,dx \leq F(b). $$
As $h \to 0$, we also have $a,b \to c$, as they are both contained in the interval $[c,c+h]$ which shrinks to a point. By continuity of $F$ we also have $F(a),F(b) \to F(c)$. Hence, by taking the limit of the three expressions above (and by squeeze theorem) we have that $$ \lim_{h \to 0}\frac1h \int_c^{c+h}F(x)\,dx=F(c), $$ which is what we wanted to prove.
As suggested by B.S. Thompson a shorter proof can be obtained using the Mean Value Theorem for Integrals. Indeed, by such theorem, we have that there exists $a \in [c,c+h]$ such that $\int_c^{c+h}F(x)\,dx=F(a)(c+h-c)=F(a)h$, that is $$ \frac1h \int_c^{c+h}F(x)\,dx=F(a). $$ Reasoning as above, when $h \to 0$, we have that $a \to c$, so by continuity of $F$ at $c$, and by taking the limits we obtain again $\lim_{h \to 0}\frac1h \int_c^{c+h}F(x)\,dx=F(c)$.