I found the following sequence in an answer of a different question, see https://math.stackexchange.com/a/1919759/579544
Consider the sequence of functions $g_n : [0,1] \to \mathbb{R}$ defined by $$ g_n(x) = \begin{cases} n - n^2x & \text{if } 0 < x < 1/n \\ 0 & \text{otherwise}. \end{cases} $$
How do I show that it converges pointwise to the zero-function?
Can you please give me a hint?
If you post the complete answer then please consider to hide it!
Simply look pointwise : for each $x > 0$, $$n > \frac 1x \implies x > \frac 1n \implies f_n(x) = 0$$. Thus for large enough $n$, $f_n(x) = 0$ so the limit is $0$. Of course, if $x =0$ then $f_n(x) = 0$ for all $n$ by definition so $f_n$ identically converges to $0$. But the lack of integral covergence ($\int_0^1 f_n(x)dx = \frac 12$ for all $n$ so the integrals don't converge) shows that the convergence is not uniform.
Proof of integral of $g_n$ : consider $g_n$ restricted to $[0,\frac 1n)$ and to $[\frac 1n,1]$. On these, apply FTC. Note that $g_n$ is discontinuous at only one point, hence Riemann integrable : it is an easy exercise to show that the integral is the sum over the two parts now. But the first is always $\frac 12$, the second always $0$.