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When trying to show this I thought of the typewriter sequence but I couldn't find a reference for a full proof, just partial statements. Here is my attempt.

We are working with the lebesgue measure on $[0,1]$. Define $$ f_n(x) = \mathbb 1_{\left[\frac{n-2^{k}}{2^k}, \frac{n-2^{k}+1}{2^k}\right]} \text{, where } 2^k \leqslant n < 2^{k+1}.$$ Result is clear for $x=0,1$. Consider all other points. Let $$\begin{align} x=0.x_1 x_2 x_3 \ldots \end{align}$$ be its binary expansion. Then define \begin{align} n_x^r = \sum_{i=1}^r x_i 2^{r-i} + 2^r\end{align} For this we have $$ k(n_x^r)= r. $$ Therefore the interval becomes $$ [S_r,S_r+\frac{1}{2^r}], $$ where $S_r$ is the $r$th partial sum from the binary expansion. By its convergence, there must exist infinitely many $r$ such that $x$ is in the interval. This establishes that there exists a subsequence such that $f_n (x) \rightarrow 1$. Now taking $n$ to be powers of $2$ shows that there exists a subsequence with $f_n(x) \rightarrow 0$.

Question: Is this correct?

Maths Wizzard
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    The example is correct indeed! It's even better than what you're trying to show, in fact - it shows convergence in $L^p$ for $p < \infty$ does not imply pointwise convergence. Relevant links: https://math.stackexchange.com/q/1281421/473276, https://math.stackexchange.com/q/138043/473276. I think your justification may be a bit overkill - you can probably just say "it's clear there's no pointwise convergence". It follows from the fact that the intervals cover $[0,1]$ infinitely often and their widths go to $0$, for example (small quibble really, your answer is fine!) – Izaak van Dongen Mar 03 '24 at 15:45
  • Hi, thanks for the response. If you wish you can post this as an answer and I will gladly accept it. – Maths Wizzard Mar 07 '24 at 00:04

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Yes, this is correct. I will expand a bit on my comment so I have something to say.

In general, if $a_n$ is an increasing sequence of numbers such that $a_n \to \infty$ and $a_{n + 1} - a_n \to 0$ (eg the harmonic numbers, $\log n$, $\sqrt n$, etc), then the sequence of indicator functions of the sets $[a_n, a_{n + 1}) \bmod 1$ is an example of a sequence which does not converge pointwise at any point in $[0, 1)$, while converging to $0$ in other senses.

This follows from that fact that for any $x \in [0, 1)$, there will be infinitely many $n$ with $x \in [a_n, a_{n + 1}) \bmod 1$, because $\bigcup_n [a_n, a_{n + 1})$ is a terminal segment of $\Bbb R$. Since $a_{n + 1} - a_n \to 0$, we also have that for sufficiently large $n$, whenever $x \in [a_n, a_{n + 1}) \bmod 1$ it's automatic that $x \notin [a_{n + 1}, a_{n + 2}) \bmod 1$. So our sequence evaluated at $x$ has infinitely many $0$s and infinitely many $1$s and hence does not converge.

Then to produce an example in $[0, 1]$ you can just make an example on $[0, 2)$ and restrict it, for example. Your example is more or less a special case of this one. (I am not suggesting that this would be a better answer to write in an exam. But to me this is the essential abstracted reason that your construction is correct).

(Sorry this took me so long! Better late than never...)

Izaak van Dongen
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