I have the following sequence of functions;
$f_n(x) = \begin{cases} 4n^2x & \text{if $0 \leq x \leq \frac{1}{2n} $} \\ -4n^2x+4n& \text{if $\frac{1}{2n} \leq x \leq \frac{1}{n} $} \\ 0 & \text{otherwise} \end{cases}$.
I want to show that $\lim_{n\rightarrow \infty} f_n(x) = 0$, and that its convergence on $[0,1]$ is not uniform. I am given the result that the limit of a sequence of riemann integrable functions need not be riemann integrable.
Note that as $n \to \infty$, the interval $\left[ 0, \dfrac 1 n \right]$ on which the graph of $f_n$ looks like a tent shrinks down to a point, while its complementary $\left[ \dfrac 1 n, 1 \right]$ on which $f_n$ is $0$ expands to cover the whole of $[0,1]$, therefore $f = \lim f_n = 0$ (the constant function $0$).
In general, to say that $f_n$ converges uniformly to $f$ on some set $A$ means that $\lim \limits _n \sup \limits _{x \in A} |f_n (x) - f(x)| = 0$. In our case this gives $\lim \limits _n \sup \limits _{x \in [0,1]} f_n (x) = 0$ (because $f = 0$ and $f_n \ge 0$). But what is $\sup \limits _{x \in [0,1]} f_n (x)$? Well, it is the value at the "top of the tent", which is $2n$ (and it is reached in the point $x = \dfrac 1 {2n}$, but this is not important here). Since $2n \not\to 0$, then the convergence is not uniform. (Note that this has nothing to do with integrability, the presence of which in this problem seems to me to be a "red herring".)