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?
