# Convergence a.s , Convergence in probability and convergence in mean

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??

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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.

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