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I'm having some difficulty with the following question:

Let $(f_{n}(x))$ be a uniformly convergent functions sequence in $(a,b)$ (where b can be $\infty$) such that $(f_{n}(x)) \to f(x)$. Suppose that for almost all $n$, the limit $\displaystyle \lim_{x\to b^{-}}f_{n}(x)$ exists.

Prove that $\displaystyle \lim_{x\to b^{-}}f(x)$ exists and $\displaystyle \lim_{n\to \infty}\left (\lim_{x\to b^{-}}f_{n}(x)\right )=\lim_{x\to b^{-}}f(x)$

I tried a method that I see often in theorems regarding uniform convergence:

First, let $\displaystyle \left ( a_{n}=\lim_{x\to b^{-}}f_{n}(x) \right )_{n\geq N}$. Such an $N$ exists.

We want to show that $\forall \epsilon,\ \exists \delta,\ \forall x \in (b-\delta, b),\ |f(x)-L|\lt \epsilon$.

We can write $|f(x)-L|=|f(x)-f_{n}(x)+f_{n}(x)-a_{n}+a_{n}-L|$, then:

$|f(x)-L|\lt\overbrace{|f(x)-f_{n}(x)|}^{\lt \epsilon / 3}+\overbrace{|f_{n}(x)-a_{n}|}^{\lt \epsilon \lt 3}+\overbrace{|a_{n}-L|}^{\text{?}}$

But I don't know how to deal with $|a_{n}-L|$. Had $a_{n}\to L$, (which is in fact the second part of the question) then I could make sure it's no more than $\epsilon / 3$, but I don't know if $a_{n}$ converges yet and if it converges to $L$.

Note: couldn't think of a better title, if anyone does, feel free to modify it.

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    @Trevor: Yes, it is supposed to be. One has to show that this exists, and that $L$ is also $\lim_{n\to\infty}a_n$.2011-01-17

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Right, the trouble is that you don't have an expression up front for $L$, so you're going to have a hard time showing that $f(x)$ converges to it (as $x \to b^-$). A good approach in this case is to use completeness of $\mathbb{R}$; then all you have to do is show that $f(x)$ is Cauchy as $x \to b^-$, and it follows that it converges to something, which you can then call $L$.

So try to show the following: for all $\epsilon > 0$, there exists $\delta > 0$ such that for all $x,y \in (b, b+\delta)$ we have $|f(x) - f(y)| < \epsilon$. By completeness, it will follow that $\lim_{x \to b^-} f(x)$ exists, and you can call that number $L$. (If you think of completeness in terms of Cauchy sequences, then note that the statement implies that for any sequence $x_n \downarrow b$, we have that $f(x_n)$ is a Cauchy sequence, hence convergent to some number $L$, and that $L$ is the same for all such sequences $x_n$.) It should not be hard then to show that $a_n \to L$ as well.

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    I ended up writing it like this: $|a_{n}-L|\leq |a_{n}-f_{n}(x)|+|f_{n}(x)-f(x)|+|f(x)-L|$ and using similar arguments from before, got the desired result. Did you have something simpler in mind?2011-01-21