integrating derivatives How can I show this?
Let $f$ be continuous on $[a,b]$ and differentiable almost everywhere on $(a,b)$. Suppose
there is a nonnegative function $g$ that is integrable over $[a,b]$ and $$|Diff_{\frac{1}{n}}f|\leq g$$ almost everywhere on $[a,b]$ for all $n$. Show that $$\int_a^b f'=f(b)-f(a),$$
where $$Diff_hf=\frac{f(x+h)-f(x)}{h}.$$
This is actually problem number 51 (Chapter 6) in Royden's Real Analysis 4th Ed.
Thanks!
 A: Let me use the notation $(\Delta_h f) (x) = \frac{f(x+h)-f(x)}{h}$, and note that for $h >0$, $\Delta_h f$ is defined on $[a,b-h]$.
Then 
\begin{eqnarray}
\int_a^{b-h} \Delta_h f &=& \frac{1}{h} \int_a^{b-h} (f(x+h)-f(x)) dx \\
&=& \frac{1}{h} \left( \int_a^{b-h} f(x+h) dx - \int_a^{b-h} f(x) dx\right) \\
&=& \frac{1}{h} \left( \int_{a+h}^{b} f(x) dx - \int_a^{b-h} f(x) dx\right) \\
&=& \frac{1}{h} \left( \int_{b-h}^{b} f(x) dx - \int_a^{a+h} f(x) dx\right) \\
\end{eqnarray}
Since $f$ is continuous, we see that $\lim_{h \downarrow 0} \int_a^{b-h} \Delta_h f = f(b)-f(a)$.
Furthermore, we see that $\lim_{h \downarrow 0} (\Delta_h f)(x) = f'(x)$ ae.
We are only a small technical detail away from using the dominated convergence theorem. Define $\phi_h(x) = \begin{cases} (\Delta_h f) (x), & x \in [a,b-h] \\ 0, & \text{otherwise}\end{cases}$. We see that $\int_a^b \phi_h = \int_a^{b-h} \Delta_h f$, and $\lim_{h \downarrow 0} \phi_h (x) = f'(x)$ ae.
We have $|\phi_{\frac{1}{n}} | \le g$ ae., hence we have $\lim_{n \to \infty} \int_a^b \phi_{\frac{1}{n}} = \int_a^b \lim_{n \to \infty} \phi_{\frac{1}{n}}(x) dx$, or in other words, $\int_a^b f' = f(b)-f(a)$.
