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I have found this sentence in a paper of F. Delbaen and W. Schachermayer with the title: A compactness principle for bounded sequences of martingales with applications. (can be found here)

On page 2, I quote: "If one passes to the case of non-reflexive Banach spaces there is—in general—no analogue to theorem 1.2 pertaining to any bounded sequence $(x_n )_{n\ge 1} $ , the main obstacle being that the unit ball fails to be weakly compact. But sometimes there are Hausdorff topologies on the unit ball of a (non-reflexive) Banach space which have some kind of compactness properties. A noteworthy example is the Banach space $ L^1 (Ω, F, P) $ and the topology of convergence in measure."

So I'm looking for a good reference for topology of convergence in measure and this property of "compactness" for $ L^1 $ in probability spaces.

Thx

math

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I may be misreading the paper, but it looks to me as if theorem 1.3 is precisely there to illustrate that sentence, as it is a version of 1.2 applicable to $L^1$. –  t.b. Dec 3 '11 at 15:13
    
In the paragraph just following theorem 1.2 they state: Note --- and this is a "Leitmotiv" of the present paper --- that, for sequences $(x_n)_{n\geq 1}$ in a vector space, passing to convex combinations usually does not cost more than passing to a subsequence. [etc.] In this sense theorem 1.3 says precisely that the topology in measure on the unit ball of $L^1$ is "sequentially compact", where passing to a subsequence is generalized to passing to certain convex combinations. –  t.b. Dec 3 '11 at 15:57
    
Another try would be the theorem stating that on uniformly integrable sets in $L^1$ the norm topology and the topology of convergence in measure coincide (and by the Dunford-Pettis theorem the former is relatively compact). –  t.b. Dec 3 '11 at 15:58
    
t.b. thx for your answer. I was just wondering, and obviously I misunderstood what they meant by this compactness property. Sorry for that. –  math Dec 3 '11 at 16:20

2 Answers 2

up vote 2 down vote accepted

So that this question has an answer: t.b.'s comment suggests that the quotes passage relates to the paper's Theorem 1.3, which states:

Theorem. Given a bounded sequence $(f_n)_{n \ge 1} \in L^1(\Omega, \mathcal{F}, \mathbb{P})$ then there are convex combinations $$g_n \in \operatorname{conv}(f_n, f_{n+1}, \dots)$$ such that $(g_n)_{n \ge 1}$ converges in measure to some $g_0 \in L^1(\Omega, \mathcal{F}, \mathbb{P})$.

This is indeed "some kind of compactness property" as it guarantees convergence after passing to convex combinations.

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Thanks to turn t.b.'s comment into an answer. –  math Aug 11 '12 at 6:26

Under the right assumptions about the measure space, any sequence of measurable functions that is Cauchy in measure actually converges in measure to some target function. Here is a set of lecture notes by C. Heil that has more information. The full webpage for that course is here.

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That's completeness, not compactness. –  Nate Eldredge Jul 12 '12 at 0:53

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