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Let Let $X=[0,1]$ with the Lebesgue measure, find a sequence $\{f_n\}$ of measurable functions $f_n:X \rightarrow{ \mathbb{R} } $ such that:

  1. $f_n(x)\rightarrow{0}$ almost everywhere $x∈[0,1]$

  2. $f_n$ converge to $0$ in measure.

  3. $f_n$ not converge weakly to $0$ in $L^p([0,1])$ for any $p$, $1≤p< \infty$

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closed as off-topic by Davide Giraudo, user91500, JonMark Perry, R_D, Daniel W. Farlow Jul 30 '16 at 13:18

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    $\begingroup$ What is a necessary condition for a sequence to be weakly convergent? $\endgroup$ – Daniel Fischer Nov 5 '14 at 1:16
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Let $$ f_n(x)= \begin{cases} n^2&\text{for $0\leq x\leq\frac1n$},\\ 0&\text{for $\frac1n< x\leq1$}. \end{cases} $$ The sequence is unbounded, hence not weakly convergent, but convergent in the sense of 1 and 2.

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  • $\begingroup$ Of note: One can verify that the sequence does not converge weakly to $0$ directly. Integrate against $\chi_{[0,1]}\in L_P^*$. $\endgroup$ – David Mitra Nov 5 '14 at 15:38

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