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

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.
