I am trying to solve the following question
True or False? If $(f_n):[0, 1] \to [0, 1]$ is a sequence of continuous functions which converges to $f$ pointwise, then $f$ is Riemann integrable and $\int_0^1{f_n} \to \int_0^1f.$
With help from the comments I found this counterexample, but I am hoping there is a simpler one.
If we replace the Riemann integrals by Lebesgue integrals, then the result is true by the Dominated Convergence Theorem. This implies that if $f$ is Riemann Integrable, then indeed $\int_0^1{f_n} \to \int_0^1f.$ So in looking for a counterexample, we should try to find one where $f$ is not Riemann integrable.
Thank you very much for any help.
The classic counterexample is the following : $\mathbf{1}_{\mathbb{Q}}(x)=\lim\limits_{n\rightarrow +\infty}\lim\limits_{m\rightarrow +\infty}\cos(n!\pi x)^{2m}$. Let $f_n(x)=\lim\limits_{m\rightarrow +\infty}\cos(n!\pi x)^{2m}$ (which exists because it is the limit of a positive decreasing sequence), then either there exists $n_0$ such that $f_{n_0}$ is not Riemann-integrable which forms a counterexample because $x\mapsto\cos(n! \pi x)^{2m}$ is Riemann-integrable for all $m$, either the $f_n$ are all Riemann-integrable, but since $\mathbf{1}_{\mathbb{Q}}$ is not Riemann-integrable and $\mathbf{1}_{\mathbb{Q}}(x)=\lim\limits_{n\rightarrow +\infty}f_n(x)$, then this is a counterexample.