# 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?

• 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. – Michael Burr Jan 28 '17 at 1:28
• The condition you are imposing is equivalent to convergence $f_n\to f$ in measure. But the latter does not imply convergence a.e. – Moishe Kohan Jan 28 '17 at 1:41
• 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? – MaudPieTheRocktorate Jan 28 '17 at 1:42

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).