# When does convergence in measure imply convergence almost everywhere?

I know that convergence almost everywhere implies convergences in measure(sometime it said locally). I also know that convergence in measure implies the existence of subsequence which is convergent almost everywhere.

However I wonder if there are some additional conditions from which the convergence in measure will imply convergence almost everywhere (for a whole sequence, not just subsequence).

Please don't mention discreteness. Let's just take real line for example, and a sequence of functions on it.

Thank you, for any help.

• monotone convergence ? – Hua Feb 20 '17 at 6:05
• Convergence a.e do not implies in general convergence in measure. This happens, for example, when the space is finite. And I never seen such sufficient condition for what you want. What you purpose with this? – Filburt Feb 20 '17 at 13:36

If the convergence in measure is rapid enough, then a.e. convergence will follow. Suppose $(f_n)$ converges in measure on the measure space $(E,\mathcal E,\mu)$. If $\mu(\{x\in E: |f_n(x)-f(x)|>\epsilon\})$ converges to $0$ so rapidly that $\sum_n\mu(\{x\in E: |f_n(x)-f(x)|>\epsilon\})<\infty$ for each $\epsilon>0$ then $\mu(\{x\in E: \limsup_n|f_n(x)-f(x)|>\epsilon\})=0$ for each $\epsilon>0$ (Borel-Cantelli lemma), so $f_n\to f$ $\mu$-a.e.