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Can I get an example of a sequence of functions $\{f_n\}$ in $L^1[0,1]$ which converges in measure such that $\lim_{n\to \infty} f_n(x)$ does not exist for every $x$?

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up vote 1 down vote accepted

given $n \in \mathbb N$, let $k_n = \lfloor \log_2 n\rfloor$ and $\ell_n := n - 2^{k_n}$. Set $f_n = \chi_{[\ell_n 2^{-k_n}, (\ell_n + 1)2^{-k_n}]}$. For $\epsilon \in (0,1)$ we have $\lambda\{f_n > \epsilon\} = 2^{-k_n} \to 0$, so $f_n \to 0$ in measure. Given $x \in [0,1]$, then for each $k \ge 3$ there are $\ell, \ell' \in \mathbb N$ with $\ell 2^{-k} \le x < (\ell+1)2^{-k}$ and $x \not\in [\ell'2^{-k}, (\ell'+1)2^{-k}]$. Hence $f_{2^k+\ell}(x) = 1$ and $f_{2^{k} + \ell'}(x) = 0$. So $f_n(x)$ is 0 for infinitely many $n$ and also $f_n(x) = 0$ for infinitely many $n$. So $\bigl(f_n(x)\bigr)$ does not converge.

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