# Weak convergence implying a.s convergence of a subsequence

Given $(f_n)$ a sequence in $L^{\infty}(\mu)$ which weakly converges to $f\in L^{\infty}(\mu),$ i.e. convergence in $\sigma(L^{\infty}, L^1),$ where $\mu$ is a probability measure. Is it true that we can extract a subsequence $(f_{n_k})$ converging almost surely to $f$ ?

## migrated from mathoverflow.netDec 4 '13 at 4:42

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• no, because $f_n$ can be e.g. a sequence of characteristic functions, $w^*$ converging to the constant $1/2$ (e.g. w.r.to the Lebesgue measure on $[0,1]$) – Pietro Majer Dec 3 '13 at 20:07
• Thank you. But what is the role of $w^*$ here? By weak convergence, it means $E[f_ng]\to E fg$ for all $g\in L^1.$ – Antoine Tran Dec 3 '13 at 20:23
• @AntoineTran: Your "weak convergence" is also known as weak-* (or w-* for short) convergence in $L^\infty = (L^1)^*$. – Nate Eldredge Dec 3 '13 at 23:13