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It is true that given any sequence of numbers, there is a monotone subsequence within. I thought obtaining a monotone subsequence out of a sequence of functions might be too much to ask. But given a sequence of $L^1$ functions, could we say that it admits a monotone subsequence?

EDIT: My motive for asking this was to decide the following problem. Given a sequence of non-negative measurable functions $\{f_n\}$, it is known that for all measurable sets $E$, $\int_Ef_n\to 0$. Could we conclude from here that $f_n\to 0$?

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Functions may be pairwise incomparable: consider, say, indicators of disjoint sets. –  Alexander Shamov Aug 21 '12 at 20:46
Regarding what you are trying to prove: the hypothesis is more or less the same as saying that the sequence converges weakly in $L^1$ —now $L^1[0,1]$ does not have the Schur property, so there are weakly convergent sequences which do not converge... –  Mariano Suárez-Alvarez Aug 21 '12 at 21:11

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Certainly not, since functions are not totally ordered. For example, if we let $f_n$ be the indicator function of $(n,n+1)$ then each $f_n\in L^1$ but none of the functions are comparable.

What you are trying to prove is however true, at least in the sense of $L^1$ convergence. Let $E$ be the whole set, and note that $$\|f_n\|_1=\left|\int_E f_nd\mu\right|=\int_Ef_nd\mu\to0$$ so $f_n\to 0$ in $L^1$, since $\|\cdot\|_1$ is a norm on the $L^1$ space, which consists of equivalence classes of functions which differ on sets of measure $0$.

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$L^1$-convergence does not imply a.e. convergence: consider, for example, indicators of $[0,1],[0,1/2],[1/2,1],\dots,[0,1/n],[1/n,2/n],\dots,[(n-1)/n,1],\dots$. They do not converge in any point. However, any subsequence contains an a.e. converging subsequence. –  Alexander Shamov Aug 21 '12 at 21:08
@AlexanderShamov You are correct, I've fixed my post. –  Alex Becker Aug 21 '12 at 21:13

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