Given $\varepsilon>0,$ one has infinitely many members of the sequence within the set $[1,1+\varepsilon/2],$ and only finitely many members of the sequence outside of that interval but still within the interval $[1,2].$ Each of those finitely many is within an open interval whose length is so small that the sum of all of those finitely many lengths does not exceed $\varepsilon/2.$ Use a partition of the interval $[1,2]$ whose points are the endpoints of those intervals. Then the upper sum is no more than $\varepsilon.$ Thus no matter how small $\varepsilon>0$ is, there is a partition whose upper sum is not more than $\varepsilon.$
We can use the Riemann criterion to prove that the uniform limit $f$ of a sequence of Riemann integrable functions $(f_n)_n$ is also Riemann integrable.
By uniform convergence, for all $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ we have
$$-\frac{\epsilon}{3(b-a)} < f(x) - f_n(x) < \frac{\epsilon}{3(b-a)}$$
Let $P: a = x_0 < x_1 < \ldots < x_n = b$ be a partition. Since $f(x) = f(x) - f_n(x) + f_n(x),$ it follows that on any partition subinterval $I$,
$$\sup_I f(x) \leqslant \sup_I(f(x) - f_n(x)) + \sup_I f_n(x) < \frac{\epsilon}{3(b-a)}+ \sup_I f_n(x), \\ \inf_I f(x) \geqslant \inf_I(f(x) - f_n(x)) + \inf_I f_n(x) > -\frac{\epsilon}{3(b-a)}+ \inf_I f_n(x).$$
Thus, $ \inf_I f_n(x)- \frac{\epsilon}{3(b-a)} <\inf_I f(x) \leqslant \sup_I f(x) < \sup_I f_n(x)+ \frac{\epsilon}{3(b-a)}. $
Summing over all partition subintervals we get for upper and lower Darboux sums,
$$U(f,P) < \frac{\epsilon}{3} + U(f_n,P), \quad -L(f,P) < \frac{\epsilon}{3} - L(f_n,P),$$
and, 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 it follows that $U(f,P) - L(f,P) < \epsilon$ proving that $f$ is Riemann integrable.
Now you should be able to prove on your own that the limit of the sequence of integrals is the integral of the limit function by considering that $|f_n(x) - f(x)| \to 0$ uniformly for all $x \in [a,b]$.
Best Answer
It is a well known result that if a sequence of continuous functions converges uniformly, its limit function is continuous . Another well known result is that any continuous function is Riemann integrable in an interval $[a,b]$. For the second part, note that
$|\int_{a}^{b} f_n\, dx - \int_{a}^{b} f \, dx| = |\int_{a}^{b} (f_n-f)\, dx|\leq \int_a^b |f_n - f|dx \leq (b-a)\max_{[a,b]}|f_n -f| $
which goes to zero as $n$ approaches to infinity