# Question about Lemma 5.9 in Royden “Real Analysis” 3rd, pages 106 - 107

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.

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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 $\lim_{h \to 0}\frac1h \int_c^{c+h}F(x)\,dx=F(c)$.
• Your proof obtains this from properties of continuous functions (assumed to be continuous on all of $[c,c+h]$). One can as easily proof it simply from the definition of continuity at $c$ (along with the same elementary manipulations of integrals). I would rather obtain it from the first mean-value theorem for integrals, since students should be encouraged to remember it for future use. – B. S. Thomson Jan 11 '16 at 17:53
• Nice. This game is more about methods than just finding one proof that works. Here is one more: if $G$ is an indefinite integral for the continuous function $F$ then $G'(c)=F(c)$ of course. But $$\frac 1h \int_c^{c+h} F(x)\,dx =\frac{G(c+h)-G(c)}{h} \to G'(c)=F(c).$$ I can't think of any more until I get some coffee. – B. S. Thomson Jan 11 '16 at 18:33