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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.

kolobokish
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  • monotone convergence ? – Hua Feb 20 '17 at 06:05
  • 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

2 Answers2

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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.

John Dawkins
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Let $f,f_n: X \rightarrow \mathbb{R}$ be measurable and $f_n \rightarrow f$ in measure $\mu$. If $f_n$ is monotone increasing then $f_n \rightarrow f$ almost everywhere.

Sketch: By basic theorems $f_n$ has a subsequence $f_{n_k}$ such that $f_{n_k}\rightarrow f$ almost everywhere. If $f_n \not\rightarrow f$ almost everywhere, by monotonicity there is a set $A \subseteq X$ with $\mu(A)>0$ such that for all $x \in A$ we have $f_n(x) \rightarrow \infty$. Again by monotonicity, for every $x \in A$ and every subsequence $f_{n_i}$ we have $f_{n_i}(x) \rightarrow \infty$. However, this would imply that there is a non-null set $A$ such that $f_{n_k} \not \rightarrow f$ on $A$. This contradicts the fact that $f_{n_k}\rightarrow f$ almost everywhere. Thus $f_n \rightarrow f$ almost everywhere.

Christopher Rose
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