Showing that the limits are not interchangeable in $\lim\limits_{n \to \infty} \int_{0}^{1} f_{n}(x)$

Question

The problem : Let $f_{n}:[0,1] \to \mathbb{R}$ be defined for $n \ge 1$ by
$$ f_{n}(x):= \begin{cases} 2n^{2} x & \text{for} & 0 \le x \le \frac{1}{2n},\\ -2n^{2} (x - \frac{1}{n}) & \text{for} & \frac{1}{2n} \le x \le \frac{1}{n},\\ 0 & \text{for} & \frac{1}{n} \le x \le 1 \end{cases} $$ [Q] : It is required to show that $ \lim\limits_{n \to \infty} \int_{0}^{1} f_{n}(x) \ne \int_{0}^{1} \lim\limits_{n \to \infty} f_{n}(x)$
My attempt :
(i) $ \lim\limits_{n \to \infty} \int_{0}^{1} f_{n}(x) = \lim\limits_{n \to \infty} (\frac{1}{2}) = \frac{1}{2} $

(ii) Here I need to show that $f_{n} \to 0$ i.e., $\lim\limits_{n \to \infty}f_{n}(x) = f(x)=0 \;,\forall x\in [0,1]$
I have an intuition that as $n \to \infty$, $\Rightarrow \frac{1}{n} \to 0$, and $x \notin \mathbin{]}0,\frac{1}{n}]$



i.e., as $n$ approaches infinity more and more values of $x$ get knocked out of $[0,\frac{1}{n}]$ and the function gets narrower. but I can't rigorously formalize this.
I would appreciate your insight regarding this part.

Answer 1

The key is to fix $x\in [0,1]$. If $x=0$ then, clearly $f_n(0)=0$ and the limit is $0$.

Now fix $x$ at $0<x_0<1$. Then for any $n>1/x_0$, we see that $f_n(x_0)=0$ and the limit is $0$.

And that is all we need to show.