# 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?

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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.

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Could you clarify the meaning of $f_n(x)$ (I am just curios as to the notation as a I have seen it elsewhere, but never understood it since I do not know the definition). – E.O. Feb 4 '12 at 0:51
It can have two meanings: 1. $f_n(x) = g(n,x)$ where $n$ runs over $\mathbb{N}$ and $x$ is real-valued. 2. $f_n(x) = \sum_{k=v}^n g(k,x)$ which is a sum of functions. Again $k$ and $n$ run over $\mathbb{N}$ and $x$ is real-valued. – Pedro Tamaroff Feb 4 '12 at 1:38