# reference for “compactness” coming from topology of convergence in measure

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

• 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

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.

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

If it's any help, convergence in measure is a metrizable criterion. For the case of a finite measure space, see Exercise 2.32 in Folland, "Real Analysis", 2nd. ed., or for a general measure space see Exercise 3.22 in my lecture notes here.

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.

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