I found the following example of a sequence of Riemann integrable functions $f_n$ whose pointwise limit function is still Riemann integrable, but $\int_0^1 f(x)dx \neq \lim_{n \to \infty} \int_0^1 f_n(x)dx$. I was wondering how we can prove the pointwise convergence rigorously.
Let $f_n:[0,1] \to \mathbb{R}$ be defined by $f_n(x) = 2n$ when $x \in \left[\frac{1}{2n}, \frac{1}{n}\right]$ and $f_n(x) = 0$ elsewhere. This sequence is supposed to converge pointwise to $0$. However, I'm having some trouble proving the fact.
For $x \neq 0$ and $\epsilon > 0$ there exists a $n_0 \in \mathbb{N}$, such that $n_0 > 1/x$ by the Archimedean property of $\mathbb{R}$. It follows that $1/n_0 < x$ and thus $x \notin \left[\frac{1}{2n_0},\frac{1}{n_0}\right]$ yielding $\left|f_n(x)\right| = 0$ for all $n \geq n_0$.
But what about $x = 0$ ? Both $\frac{1}{2n}$ and $\frac{1}{n}$ converge to $0$ for $n \to \infty$. I don't see how the sequence converges to the zero function at $x = 0$:
