Proof of 5.9 Corollary from Bartle (Integrable Function) 



This is form p. 46-47 from the Bartle's text. I don't understand one step in this proof. At the end of the proof, it says that "this integrand is dominated by $g(x)$". We know that $|\frac{\partial{f}}{\partial{t}} (x,t) | \le g(x)$, but does this imply that $|\frac{f(x,t_n) - f(x, t)}{t_n - t}| \le g(x)$?  I think that this is not necessarily true unless if we let n goes to infinity. If not true, we cannot use Dominated Convergence Theorem.
I appreciate if you explain the last step of this proof.
(By the way, $F(t) = \int_X f(x, t) d\mu(x)$.)
 A: Since $t \mapsto f(x,t)$ is differentiable for all $(x,t) \in X \times [a,b]$ and $t_n,t \in [a,b]$ we can apply the mean value theorem which guarantees the existence of a point $\xi$ between $t_n$ and $t$ such that
$$\left|\frac{f(x,t_n) - f(x,t)}{t_n - t} \right| = \left|\frac{\partial f}{\partial t}(x,\xi) \right| \leqslant g(x)$$
That $\xi$ may depend upon $x$ and $n$ is immaterial since the dominating  function $g$ applies for any $t \in [a,b]$.
A: Looks like it's been a while, but I'm amending the main ideas left out in the proof to aid the future reader.
First note that every member in the sequence of functions defined by 
$$\frac{f(x,t_n)-f(x,t)}{t_n - t} = \left(\frac{1}{t_n-t}\right)f(x,t_n)-\left(\frac{1}{t_n-t}\right)f(x,t)$$
is $\mathbf{X}$-measurable because it is a linear combination, and $x\rightarrow f(x,t)$ is assumed to be $\mathbf{X}$-measurable for each $t\in[a,b]$ (Bartle, p. 45). Therefore the limit $\partial f (x,t)/\partial t $ will also be $\mathbf{X}$-measurable (Corollary 2.10, p. 12). 
We now must show that this sequence of functions is integrable, and we do that by showing $x\rightarrow f(x,t)$ is integrable. To do that, note that we assumed $\partial f/\partial t$ exists, or $f$ is differentiable with respect to $t$, on $X\times[a,b]$. This is a sufficient condition to invoke the Mean Value Theorem which guarantees the existence of an $s_1 \in (t_0,t)$ such that:
$$f(x,t)-f(x,t_0) = (t-t_0)\frac{\partial f}{\partial t} (x,s_1)$$
Now, using the Triangle inequality, and the fact that $|\partial f/\partial t|$ is bounded by $g(x)$ we get that:
$$|f(x,t)|=|f(x,t_0) + (t-t_0)\frac{\partial f}{\partial t} (x,s_1)| \leq |f(x,t_0)|+|t-t_0|g(x)$$
Now, since we have assumed in the statement of the corollary above that $x\rightarrow f(x,t_0)$ is integrable, that $x\rightarrow g(x)$ is integrable, and (on p. 45) that $x\rightarrow f(x,t)$ is measurable, we can claim that $x\rightarrow f(x,t)$ is also integrable (Corollary 5.4, p 43). 
At this point we have a sequence of integrable functions (by virtue that $x\rightarrow f(x,t)$ is integrable for each $t\in [a,b]$, as we have just proven) that converges to an $\mathbf{X}$-measurable $\partial f (x,t)/\partial t $, and that every term in the sequence of functions is bounded by an integrable function $g$ (by the Mean Value Theorem, see RRL's answer). We can now finally apply Lebesgue's Dominated Convergence Theorem to get:
$$F'(t) = \frac{d}{dt}\int f(x,t)d\mu(x) = \lim_{n\rightarrow \infty}\frac{F(t_n)-F(t)}{t_n-t} = \lim_{n\rightarrow \infty} \int\frac{f(x,t_n)-f(x,t)}{t_n-t}d\mu(x)
= \int\frac{\partial f}{\partial t}(x,t)d\mu(x)$$
$$\blacksquare$$
