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Let $f_i: [0, 1] \to \mathbb R$ be functions in $L^1 \cap L^\infty$ with $\sup_i \|f_i\|_{L^\infty} < M$ for some $M > 0$.

Suppose $f_i$ converge weakly in $L^1$ to some $L^1$ function $f$ - that is, $\int f_i g \ d\mu \to \int fg \ d\mu$ for all $g \in L^\infty$.

Question: Does there exist a subsequence $f_{n_k}$, and measure preserving bijections $T_k: [0, 1] \to [0, 1]$ such that $f_{n_k} \circ T_k$ converges strongly in $L^1?$ (Not necessarily to $f$)

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    $\begingroup$ What do you mean by weak-$*$ convergence in $L^1$? $\endgroup$
    – Jochen Glueck
    Commented Aug 6, 2021 at 6:53
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    $\begingroup$ Thanks for the clarification! This is actually called weak (rather than weak-$*$) convergence. $\endgroup$
    – Jochen Glueck
    Commented Aug 6, 2021 at 8:27
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    $\begingroup$ Ah, sorry my bad. I will modify it, and also write the definition in the original post. $\endgroup$
    – Nate River
    Commented Aug 6, 2021 at 8:44
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    $\begingroup$ Isn't any set of decreasing uniformly bounded functions pre-compact in $L^1$? $\endgroup$
    – fedja
    Commented Aug 6, 2021 at 12:55
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    $\begingroup$ @fedja Oh, I think that is true - in that case we just rearrange all the $f_i$ to be decreasing and we are done then. $\endgroup$
    – Nate River
    Commented Aug 6, 2021 at 15:43

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