Let $f_n(x)$ and $f(x)$ be continuous functions on $[0, 1]$ such that $\lim_{n\to\infty} f_n(x) = f(x)$ for all $x \in [0, 1]$. Answer each of the following questions. If your answer is “yes”, then provide an explanation. If your answer is “no”, then give a counterexample.
(a) Can we conclude that $\lim_{n\to\infty}\int_{0}^{1}f_n(x)\,dx = \int_{0}^{1}f(x)\,dx$?
(b) If in addition we assume $|{f_n(x)}|\leq 2017$ for all n and for all $x \in [0, 1]$, can we conclude that $\lim_{n\to\infty}\int_{0}^{1}f_n(x)\,dx = \int_{0}^{1}f(x)\,dx$?
My attempt:
I could not think of any counterexample for part (a).
For part (b), I showed that ${f_n}$ is equicontinuous which along with uniform boundedness gives me that ${f_n}$ has a uniformly convergent subsequence. (Arzela Ascoli). Hence $\lim_{n\to\infty}\int_{0}^{1}f_n(x)\,dx= \lim_{k\to\infty}\int_{0}^{1}f_{n_k}(x)\,dx = \int_{0}^{1}f(x)\,dx$. Is that right?
Edit: I have made a mistake while proving equicontinuity.
I know that monotonicity + continuity + pointwise convergence on [0,1] ensures uniform convergence. Is it true that uniform boundedness + continuity + pointwise convergence ensures uniform convergence? How do I show that?
The question of switching limits is an important one, and with the Lebesgue integral one gets the powerful dominated convergence theorem. However, I presume you are using the Riemann integral.
For part a), I will give you a place to start. We denote the function $ \chi_S : \mathbb R \to \mathbb R $ via
\begin{equation*} \chi_S(x) = \begin{cases} 1 & x \in S \\ 0 & x \notin S \end{cases} \end{equation*}
For any subset $ S \subseteq \mathbb R $.
Let $ f_n = n \chi_{(0, \frac{1}{n})} $. Check that $ f_n(x) \to 0 $ for all $ x \in [0, 1] $, so $ f(x) = 0 $. Hint: do this first for $ x = 0 $, then for $ x \in (0, 1] $. It is easy to see that $ \int_{0}^{1} f_n(x) \,dx = 1 $ for all $ n $. Then $ f_n \to f $ pointwise, but $ \int_{0}^{1} f_n(x) \,dx \not\to 0 = \int_{0}^{1} f(x) \,dx $. These functions are not continuous, so I will leave it to you to modify these to get a continuous version of this counterexample. In fact, it is possible to find a counterexample where each function is smooth.
For part b), your proof is close, but you need more detail to fill it out. Specifically, you have shown that $ f_{n_k} $ converges uniformly to $ f $, but you have not shown this for $ f_n $. To do this, you must use the fact that $ f_n \to f $ pointwise, as this is false otherwise. Suppose then that $ f_n $ does not converge uniformly to $ f $. Then, there exists an $ \epsilon > 0 $ such that for some $ n_m \to \infty $, there exists a sequence $ x_{n_m} $ in $ [0, 1] $ such that $ |f_{n_m}(x_{n_m}) - f(x_{n_m})| \geq \epsilon $. We get this by negating the definition of uniform convergence. However, this tells us that no subsequence of $ f_{n_m} $ converges uniformly to $ f $, but Arzela Ascoli guarantees the existence of a uniformly convergent subsequence, and the pointwise convergence of $ f_n $ to $ f $ tells us that this uniformly convergent subsequence must converge to $ f $. This is a contradiction, so $ f_n \to f $ uniformly, so $ \int_0^1 f_n(x) \, dx \to \int_0^1 f(x) \, dx $.