Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anybody help me, Im trying to think of an example to show that convergence in mean does not imply convergence P a.s and also to show in the other direction, i.e convergence in P a.s does not imply convergence in mean??

share|cite|improve this question
Hint: The dominated convergence theorem tells you that an example in which a.s. convergence does not imply mean convergence must be very much unbounded. For the other direction, just take nondegenerate iid random variables. – Michael Greinecker Jun 7 '12 at 10:58

Consider $\Omega=[0,1]$ with Lebesgue measure.

  • For an integer $n$, there is an integer $k_n$ such that $1+\dots+2^{k_n}\leq n<1+\dots+2^{k_n+1}$. We define $$X_n:=2^{-k_n/2}\chi_{((n-(1+\dots+2^{k_n}))2^{-k_n},((n-(1+\dots+2^{k_n})+1)2^{-k_n})}.$$ More concretely, we have $1/4\chi_{(0,1/2)}$, $1/4\chi_{(1/2,1)}$, $1/8\chi_{(0,1/4)}$. We have convergence in $L^1$ to $0$ but not almost everywhere.

  • Take $X_n:= n\chi_{(0,1/n)}$. It converges almost everywhere to $0$ but not in mean.

However, convergence in means implies that we can extract a subsequence which converges almost everywhere.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.