Question from Stein's Fourier Analysis on sequence of integrable functions The following question is from Stein and Shakarchi's Fourier Analysis, specifically chapter 3 (Convergence of Fourier series): 
Construct a sequence of integrable functions $\{f_k\}$ on $[0, 2 \pi]$ such that 
$$\lim_{k \to \infty} \int_0^{2\pi}|f_k(\theta)|^2 d \theta = 0$$ 
but $\lim_{k \to \infty} f_k(\theta)$ fails to exist for any $\theta$. 
[Hint: Choose a sequence of intervals $I_k \subset [0, 2 \pi]$ whose lengths tend to $0$, and so that each point belongs to infinitely many of them; then let $f_k = \chi_{I_k}$.] 
I know the hint gives a general idea, but could someone spell this out for me? Specifically, I don't really get what he means by 'so that each point belongs to infinitely many of them'. And how can infinitely many points belong to the intervals $I_k$ if their lengths tend to $0$? 
There doesn't seem to be much in the way of help in the preceeding chapter, so it would be great if anyone can shed any light on it. Thanks very much in advance.  
 A: On $[0,1]$ consider the typewriter sequence:
$$\chi_{[0,1]},\, \chi_{[0,\frac12]},\,\chi_{[\frac12,1]},\,\chi_{[0,\frac13]},\, \chi_{[\frac13,\frac23]},\,\chi_{[\frac23,1]},\dots $$
A: The idea below uses the fact that $\sum_{n \in \mathbb{N}}\frac{1}{n}$ diverges. Using this hint, construct a sequence $\{I_k\}$ of intervals such that the length of each interval is $\frac{1}{k}$ and at each step, intervals are shifting to right. More precisely, below is a construction for $[0,1]$:


*

*$I_1$ is of length $1$, and covering the entire interval

*Left end point of $I_2$ will coincide with $0$. We will then recover a sequence such that $\frac{1}{2}+\frac{1}{3}+\dots + \frac{1}{n} \geq 1$ and $\frac{1}{2}+\frac{1}{3}+\dots + \frac{1}{n-1} < 1$. Such an $n$ must exist since harmonic series diverge.

*Left end point of $I_3$ will coincide with right end point of $I_2$ (e.g. in our example, $I_2 =\left[0,\frac{1}{2}\right], I_2 =\left[\frac{1}{2},\frac{5}{6}\right]$ and so on.

*Continue until $I_{n-1}$. For $I_n$, adjust it so that its right endpoint coincides with 1 (in particular, place it so that the right end point of $I_n$ coincides with 1).

*Once we reach $n$, the next stack starts. Recover a $k$ such that
\begin{equation}
\frac{1}{n+1}+\frac{1}{n+2}+\dots + \frac{1}{n+k} \geq 1
\end{equation}
and
\begin{equation}
\frac{1}{n+1}+\frac{1}{n+2}+\dots + \frac{1}{n+k-1} < 1
\end{equation}
Apply the same steps above. $I_{n+1} = \left[0,\frac{1}{n+1}\right], I_{n+2}=\left[\frac{1}{n+1},\frac{2n+3}{n^2+3n+2}\right]$ and so on. Make sure that $I_{n+k}$ does not go outside the interval.


With this construction, it is clear that every $\theta$ belongs to infinitely many of such intervals. Hence, if we let $f_k = \mathcal{X}_{I_k}$ ($\mathcal{X}$ is the indicator function) then for every fixed $\theta$, the sequence $\{f_k(\theta)\}_{k=1}^{\infty}$ has infinitely many 0's and 1's; hence cannot converge.
On the other hand, 
\begin{equation}
\frac{1}{2\pi}\int_{0}^{2\pi}|f_k(\theta)|^2 d\theta = \frac{1}{2\pi}\frac{1}{k^2} \to 0, \ \text{as} \ k \to \infty
\end{equation} 
Hence, this construction indeed works. 
