I have problems with understanding the following example:
Suppose $\left( \Omega, \mathcal{F}, \mathbb{P}\right)=\left([0,1], \mathcal{B}([0,1]) , \lambda|_{[0,1]}\right)$ and the sequence of random variables $\left(f_n\right)_{n \in \mathbb{N}}$ defined by
$f_n(\omega):= \begin{cases} n^2 2^{n+1} \omega, & \omega \in \left[0,\frac{1}{2n}\right],\\ n 2^{n+1}-n^2 2^{n+1} \omega, & \omega \in \left( \frac{1}{2n},\frac{1}{n}\right],\\ 0, & \omega \in \left(\frac{1}{n},1\right]. \end{cases}$
Then we have $\lim_{n \to \infty} f_n(\omega)=0$ for all $\omega \in [0,1]$ but $\lim_{n \to \infty} \mathbb{E} f_n = \lim_{n \to \infty} 2^{n-1}=\infty$.
First question: I don't understand why we have $\lim_{n \to \infty} f_n(\omega)=0$. Because if we take $\omega=\frac{1}{2n}$ we get $\lim_{n \to \infty} f_n\left(\frac{1}{2n} \right)= \lim_{n \to \infty}n 2^{n}=\infty$.
Second question: How can I calculate $\mathbb{E} f_n$?
Take any fixed $\omega$ with $0 \lt \omega \le 1$; then for $n \gt \frac{1}{\omega}$ you have $f_n(\omega)=0$, so $\displaystyle\lim_{n \to \infty} f_n(\omega)=0$. In addition $f_n(0)=0$ for all $n$. This means you have pointwise convergence of $f_n$ to $0$ (Wikipedia calls this "sure convergence", and it is also convergence in probability).
Meanwhile $\displaystyle \mathbb{E}[f_n]=\int_{0}^{1} f_n(\omega) \,d\omega=\int_0^{1/2n} n^2 2^{n+1} \omega \,d\omega+\int_{1/2n}^{1/n} (n 2^{n+1}-n^2 2^{n+1} \omega) \,d\omega $ which I suspect gives an increasing power of $2$ as a result.