Show that a sequence of functions that converge uniformly are Riemann integrable. What if they only converge point wise? Let $f_n$ be a sequence of Riemann integrable functions on $[a,b]$ that converge uniformly to a function f. Show that f is also Riemann integrable. What happens if $f_n$ only converges pointwise?
Considering this scenerio, show that
$$\lim_{n \to \infty} \int_{a}^{b} f_n(x) dx = \int_{a}^{b} f(x) dx$$
Not really sure how to start this problem. If they already converge to a function, uniformly, that function has to be continuous right? So then it is trivially Riemann integrable. The pointwise case I'm not sure about. And then the second part with the limits I'm not sure how to go about. Any help is appreciated!
 A: Let $\{q_n\}_{n \in \mathbb{N}}$ be the rational numbers in the interval $[0,1]$, and let us consider the functions
$$f_n(x)=
\begin{cases}
1 &\text{if}\quad x \in \{q_1, \ldots, q_n\},\\
0 &\text{otherwise.}
\end{cases}
$$
The $f_n(x)$ are Riemann-integrable but they converge to the Dirichlet function, which is not Riemann-integrable.
A: 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 $x \in [a,b]$.
