Uniform continuity allows pushing limit inside integral? Let's say I have a (uniformly) continuous functions $f:[a,b] \to \mathbb{R}$ and an arbitrary function $h:\mathbb{R}^2 \to [a,b]$ such that
$$
\lim_{t\to 0} ~h(t,x) = h(0,x) = x
$$
for all $x$.
I would like to be able to conclude that
$$
\lim_{t\to 0}\int_a^bf(h(t,x))dx = \int_a^bf(x)dx.
$$
Is pushing the limit inside the integral justified in this situation?
 A: As you state, if we define $$y_n = f_n(x) $$ and $$y = f(x) = \lim_{n \to \infty} f_n(x)$$
then it is legitimate to state $$\mathop {\lim }\limits_{n \to \infty } \int\limits_a^b {{f_n}\left( x \right)dx = } \int\limits_a^b {\mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right)dx = \int\limits_a^b {f\left( x \right)dx} } $$
if the convergence is uniform over $[a,b]$. 
I leave here a theorem I got from Piskunov's or Apostol's Calculus (don't remember which):
THEROEM Let $\sum u(x)_k$ be a series of functions, uniformly convergent in a closed interval $I$. Then if $x, \alpha\in I$ $$\int_{\alpha}^x s(t) dt = \sum \int_{\alpha}^x u_k(t) dt$$ where s(x) is the sum of the series (i.e. the limit as $n \to \infty$). This is usually stated as "an uniformly convergent series can be integrated term-wise".
Moreover, if $\sum u(x)_k$ and $\sum u'(x)_k$ are U.C. then you have that $$s'(x) = \sum u'(x)_k$$ 

EDIT: I'll add two examples of the notation $f_n(x)$
Let $f_n(x) = \tanh(nx)$. Then it is clear that the $f_n(x)$ converges to $0$ in $x=0$ and $\frac{\pi}{2}$ in $x>0$. Thus you have a sum of continuous functions that converge to a discontinuous.
Let $f_n(x) = \displaystyle  \displaystyle \sum_{k=0}^{n} \frac{xn^2}{2^n}$. Then the function converges to $y = 6 x$ (note that $x$ is not indexed). There are more complex cases you might want to look at in books and webs.
