Proving a sequence of integrable functions that converges uniformly converges to an Integrable function So I have a sequence of riemann (darboux specifically) integrable real valued functions $(f_n)$ defined on $[a,b] \subset \mathbb{R}$ that converges uniformly to some $f$. I would like a hint (or two even) on how to show this $f$ is integrable. I split this problem up into two parts, namely first proving that $f$ is bounded, and then later proving that $f$ is integrable. So for the first part I have that:
Recall that each $f_n$ is bounded on $[a,b]$, so we have that for each $n$, $f_n([a,b])$ has some supremum in $\mathbb{R}$ (call it $M$) and that:
$$
f_n(x) \leq M \text{ for any $x\in [a,b]$}
$$
A similar statement can be made about the lower bound. Fix some $\epsilon>0$. We know by the uniform convergence of $f_n$ to $f(x)$ that, there exists $N \in \mathbb{N}$ with the property that for any $n \geq N$ we have:
$$
|f_n(x) - f(x)|<\epsilon \text{ for all $x\in [a,b]$}
$$
So we see that:
$$
f(x)< f_n(x) + \epsilon \leq M +\epsilon \text{ for all $x\in [a,b]$}
$$
So $f$ is bounded.
I am not entirely sure how to proceed from here. I am fairly certain I need to satisfy the Riemann integrability condition, i.e show that for any $\epsilon$ there exists a partition $P$ of $[a,b]$ such that:
$$
U(f,P) - L(f,P) < \epsilon
$$
I have a feeling I need to pick some special epsilons and do some tricky manipulations, but I was hoping for some guidance in the right direction.
 A: HINT: Write
\begin{equation*}
\begin{aligned}
U(f,P) - L(f,P) &= U(f,P) - U(f_n,P) + U(f_n,P) - L(f_n,P) + L(f_n,P) - L(f,P) \\
&\leq \vert U(f,P) - U(f_n,P) \vert + U(f_n,P) - L(f_n,P) + \vert L(f_n,P) - L(f,P) \vert.
\end{aligned}
\end{equation*}
As an aside, it's worth noting that once you prove this, you can also use uniform continuity to prove that
$$\lim_{n\to\infty}\int_a^b f_n(x)dx = \int_a^b f(x)dx,$$
i.e. you can interchange the order of the limit and the integral.
A: Some care is required in estimating $U(f,P) - L(f,P)$.  
By uniform convergence, if $n$ is sufficiently large, then
$$-\frac{\epsilon}{3(b-a)} < f(x) - f_n(x) < \frac{\epsilon}{3(b-a)}.$$
We have
$$f(x) = f(x) - f_n(x) + f_n(x),$$
implying, on any partition subinterval $I$,
$$\sup_I f(x) \leqslant \sup_I(f(x) - f_n(x)) + \sup_I f_n(x) < \frac{\epsilon}{3(b-a)}+ \sup_I f_n(x), \\ \inf_I f(x) \geqslant \inf_I(f(x) - f_n(x)) + \inf_I f_n(x) > -\frac{\epsilon}{3(b-a)}+ \inf_I f_n(x).$$
The second chain of inequalities implies 
$$-\inf_I f(x) <  \frac{\epsilon}{3(b-a)} - \inf_I f_n(x).$$
Summing over all partition subintervals we get
$$U(f,P) < \frac{\epsilon}{3} + U(f_n,P),\\ -L(f,P) < \frac{\epsilon}{3} - L(f_n,P).$$
Hence, 
$$U(f,P) - L(f,P) < \frac{2\epsilon}{3} + U(f_n,P) - L(f_n,P).$$
Since $f_n$ is Riemann integrable, there is a partition $P$ such that $U(f_n,P) - L(f_n,P) < \epsilon/3$ and the desired result $U(f,P) - L(f,P) < \epsilon$ follows.
A: You may not like this method, but here goes!
A function is Riemann integrable on $[a,b]$ iff it's bounded, Lebesgue measurable
and continuous almost everywhere in the Lebesgue sense. Boundedness
and Lebesgue measurability are preserved under uniform limits.
As a countable union of measure zero sets has measure zero, then it
suffices to prove that $f$ is continuous at each $x\in[a,b]$
at which all the $f_n$ are continuous. The usual "$3\varepsilon$ argument"
for continuity of uniform limits works for this.
