Suppose $f_{n}$ is a sequence of functions in $L^{p}(\mathbb{R}^{d})$ such that $\|f_{n}\|_{L^{p}} \leq 1$ for all $n$ and $f_{n}(x) \rightarrow f(x)$ pointwise almost everywhere as $n \rightarrow \infty$. If I also know that $f_{n}$ converges weakly in $L^{p}$ to an $\bar{f}$, then is $\bar{f} = f$ a.e.?
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Hint : Do you know Egorov's Theorem? – Euler....IS_ALIVE Jul 15 '14 at 21:34
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See this. – David Mitra Jul 15 '14 at 21:42
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1The boundedness condition is redundant, since weakly convergent sequences are bounded. – David Mitra Jul 15 '14 at 21:44
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Actually, for $1<p<\infty$, your conditions in the first sentence imply weak convergence of $(f_n)$ to $f$. – David Mitra Jul 15 '14 at 22:01