Let $\left\lbrace f_{n} \right\rbrace$ be a sequence of bounded functions that converges uniformly to a bounded function $f$. I need to prove that: $$ \lim_{n\rightarrow \infty}\overline{\int_{a}^{b}}f_{n}(x)dx = \overline{\int_{a}^{b}}f(x)dx. $$
So far I have that, by uniform convergence and boundedness, given $\epsilon>0, \exists M\in\mathbb{R}$ s.t. $\forall n\geq M, \left| f_{n}(x) - f(x) \right| < \epsilon.$
Thus I know that $f_{n}(x)$ and $f(x)$ can be made arbitrarily close, such that all I need to do is force $f(x)$ to be within $\epsilon$ of "something else," and then force "something else" to be within $\epsilon$ of $\overline{\int_{a}^{b}}f(x)dx$.
I'm stuck on finding that "something else," though I suspect this is a typical $\frac{\epsilon}{C}$ argument using the triangle inequality.
Thanks for any hints.
For every $\varepsilon>0$, there exists an $n_0$, such that $n\ge n_0,$ implies that $|f_n(x)-f(x)|<\varepsilon$, for all $x\in [a,b]$ and $n\ge n_0$. Thus $$ f(x)-\varepsilon<f_n(x)<f(x)+\varepsilon, \tag{1} $$ for all $x\in [a,b]$ and $n\ge n_0$. Take now an arbitrary partition $P=\{a=t_0<\cdots< t_n=b\}$ of $[a,b]$. Then $(1)$ implies that $$ \sup_{x\in[t_{i-i},t_i]} f(x)-\varepsilon \le \sup_{x\in[t_{i-i},t_i]} f_n(x) \le \sup_{x\in[t_{i-i},t_i]} f(x)+\varepsilon $$ and thus \begin{align} \sum_{i=1}^n(t_i\!-\!t_{i-1})\!\!\sup_{x\in[t_{i-i},t_i]} \!f(x)\!-\!\varepsilon(b\!-\!a) \le \sum_{i=1}^n(t_i\!-\!t_{i-1})\!\!\sup_{x\in[t_{i-i},t_i]} \!f_n(x) \le \sum_{i=1}^n(t_i\!-\!t_{i-1})\!\!\sup_{x\in[t_{i-i},t_i]} \!f(x)+\varepsilon(b\!-\!a), \end{align} or $$ U(f,P)-\varepsilon(b-a)\le U(f_n,P)\le U(f,P)+\varepsilon(b-a), $$ and after that you know what to do.