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.