Every subsequence of $\{f_{n}\}$ has a subsequence convergent almost everywhere; does this imply convergence almost everywhere? $f,\ f_{n}: X \rightarrow \mathbb{R}$ are measurable functions, such that every subsequence of $\{f_{n}\}$ has a subsequence that converges to $f$ almost everywhere. Does that mean $f_{n}$ converge to $f$ almost everywhere?
 A: Not in general, no. Take $X=[0,1]$ with Lebesgue measure (or any nontrivial measure), and let $f$ be identically $0$; consider the sequence $\{f_n\}$ where each function is the characteristic function of an interval (that is, equal to $1$ on the interval and $0$ elsewhere), where the sequence of intervals looks like
$$
\textstyle
[0,1]; [0,\frac12], [\frac12,1]; [0,\frac14], [\frac14,\frac12], [\frac12,\frac34], [\frac34,1]; [0,\frac18], [\frac18,\frac14], \dots.
$$
This sequence doesn't converge pointwise anywhere. However, every subsequence $\{f_{n_m}\}$ of $\{f_n\}$ has a subsubsequence converging to $f$:


*

*Either infinitely many of the $f_{n_m}$ are identically $0$ on $[0,\frac12)$, or infinitely many of the $f_{n_m}$ are identically $0$ on $(\frac12,1]$. WLOG, the first is true; choose a subsequence $f_{n_{m_l}}$ where each term is identically $0$ on $[0,\frac12)$.

*Either infinitely many of the $f_{n_{m_l}}$ are identically $0$ on $[\frac12,\frac34)$, or infinitely many of the $f_{n_{m_l}}$ are identically $0$ on $(\frac34,1]$. WLOG, the first is true; choose a subsequence $f_{n_{m_{l_k}}}$ where each term is identically $0$ on $[\frac12,\frac34)$.

*And so on and so on; this yields an infinite nested list of subsequences. Then take the diagonal subsequence (the first function of the first subsequence, the second function of the second subsequence, and so on); this is also a subsequence of the original subsequence, and it's not hard to see that this subsequence converges almost everywhere to $f$ (perhaps even everywhere, depending on how we define these functions at the endpoints of their intervals).


The answer is yes if $X$ is a singleton set (this is a common result in analysis), and by the same diagonal argument as above, the answer is also yes if $X$ is countable (or if the measure is supported on a countable set).
