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

Question

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??

Answer 1

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.