Particular sequence in $L^2[0,1]$ that fades at $\infty$ in $L^2$ norm and whose limit is not defined $\forall$ t I am looking for a sequence $\omega_n \in L^2[0,1]$ s.t. $\omega_n \rightarrow 0$ in $L^2$, $\lim_{n\to\infty} \omega_n(t)$ is not defined $\forall \text{ } t$.
I was thinking about something involving periodic functions so that the limit is not defined, but at the same time it should satisfy the condition $\omega_n \rightarrow 0$. Any suggestions?
 A: Start from the function $\omega(x)=|x|$, defined in $[-1,1]$, and extend it periodically over $\mathbb{R}$. Then set
$$\omega_n = \omega(nx)^n ,\qquad x\in [0,1]$$
Edit: Never mind this previous example, here is a much simpler one to work with. Construct a sequence $\omega_n$ in the following way. 
\begin{align*}
\omega_1&=\chi_{[0,1/2)},\qquad \omega_2=\chi_{[1/2,1]}\\ 
\omega_3&=\chi_{[0,1/3)},\quad \omega_4=\chi_{[1/3,2/3)},\quad \omega_5=\chi_{[2/3,1]}\\
\omega_6&=\chi_{[0,1/4)},
\end{align*}
and so on. Clearly $\|\omega_n\|_2\to 0$ because the sequence $\left\{\omega_n\right\}$ consists of a sequence of characteristic functions of intervals with measures converging to $0$. Moreover, for each $x_0\in [0,1]$ and for each $n\in \mathbb{N}$ there is a $k\in \left\{0,\dots,n-1\right\}$ so that 
$$x_0\in \left[\frac{k}{n},\frac{k+1}{n}\right) $$
So that when the sequence $\left\{\omega_n\right\}$ gets to the term $\omega_{\alpha_n}$ given by the characteristic function of this interval, we will have $\omega_{\alpha_n}(x_0)=1$. On the other hand from the way the sequence is constructed we see that $\omega_{\alpha_n+1}(x_0)=0$, and therefore we have two subsequences of $\left\{\omega_n\right\}$ which are constantly equal to $1$ and $0$ on $x_0$ respectively. Hence $\lim_{n\to \infty}\omega_n(x_0)$ does not exist.
