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$f,\ f_{n}: X \rightarrow \mathbb{R}$ 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?

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  • $\begingroup$ Hint: Write down what it would mean for $f_n$ to not converge to $f$ almost everywhere. This meaning should include a subsequence. Then, relate this subsequence to the given property. $\endgroup$ – Michael Burr Jan 28 '17 at 1:28
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    $\begingroup$ The condition you are imposing is equivalent to convergence $f_n\to f$ in measure. But the latter does not imply convergence a.e. $\endgroup$ – Moishe Kohan Jan 28 '17 at 1:41
  • $\begingroup$ I don't understand. I get that $\exists\epsilon>0 \forall n\geq 1 \exists k\geq n, \mu({x\in X : |f_k(x) - f(x)| > \epsilon})>0$, but then what? $\endgroup$ – MaudPieTheRocktorate Jan 28 '17 at 1:42
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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).

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  • $\begingroup$ Why do you have to diagonalize the nested list of subsequences? It seems fine to just take the first function of each subsequence. $\endgroup$ – MaudPieTheRocktorate Jan 28 '17 at 1:48
  • $\begingroup$ It seems you're right! I'm so used to the diagonal construction, I didn't think about that. $\endgroup$ – Greg Martin Jan 28 '17 at 1:57

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