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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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    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...2012-08-21

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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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    @AlexanderShamov You are correct, I've fixed m$y$ post.2012-08-21