TITLE: Differentiating an integral using dominated convergence
QUESTION [12 upvotes]: Let's say we have
$$F(x)=\int^b_af(x,t)\ \mathrm{dt}$$
And we want to calculate $F'(x)$. Then:
$$F'(x)=\lim_{h\to0}\frac{\int_{a}^{b}f(x+h,t)\mathrm{dt}-\int_{a}^{b}f(x,t)}{h}\mathrm{dt}=\lim_{h\to0}\int_{a}^{b}\frac{f(x+h,t)-f(x,t)}{h}\mathrm{dt}\tag1$$
Now, in class we were given (without proof) the proposition that $F'(x)=\int^b_a\frac{df}{dx}(x,t)\ \mathrm{dt}$ under certain conditions, one of which is that $\frac{df}{dx}(x,t)$ be dominated by some integrable function $g(t)$.
If I could just commute the $\lim$ and the $\int$ in $(1)$, I could prove this proposition. In order to justify that, I need to find an integrable $g(t)$ such that:
$$\frac{f(x+h,t)-f(x,t)}{h}\leq g(t)$$
in a neighborhood of $0$ (right?). This seems very similary to the requirement we learned in class, but how is the boundedness of $\frac{df}{dx}(x,t)$ equivalent to the boundedness of the above? All I can see is that as $h\to0$, the difference quotient is pointwise eventually smaller than $g$. But I don't see that I can conclude that there is therefore a neighborhood of $0$ where the difference quotient is bounded by $g$.

REPLY [15 votes]: By the mean value theorem, there is, for each $t$, a $c_t(x) \in (0,1)$ with
$$\frac{f(x+h,t) - f(x,t)}{h} = \frac{\partial f}{\partial x}(x + c_t(x)\cdot h, t).$$
So if the partial derivative $\frac{\partial f}{\partial x}$ is locally uniformly (in $x$) dominated by $g$, then so are the difference quotients. And that makes the dominated convergence theorem applicable.