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Let $(\Omega,\Sigma, \mu)$ be a probability measure and $X$ a Banach space. I am interested in subsets $F\subseteq L_\infty (\mu,X)$ that satisfy these two compactness conditions:

  1. $F$ is a norm-compact subset of $L_1(\mu, X)$; and
  2. For any sequence $f_n$ in $F$ there exists a set $E\in \Sigma$, $\mu(E)=1$, such that each $L_1$-norm convergent subsequence of $f_n$ converges pointwise on $E$.

An example of such a set is any compact subset $F$ of $L_\infty (\mu,X)$. To see this suppose that $f_n$ is a sequence in $F$. There is $E\in \Sigma$, $\mu(E)=1$, such that for any $m,n$ we have $\|f_m(\omega) - f_n(\omega)\|\leq \|f_m-f_n\|_\infty$ for all $\omega\in E$. Now if $f_m$ is an $L_1$-convergent subsequence of $f_n$, then it must be converging in $L_\infty$, and converging pointwise on $E$.

More generally, the compactness condition is satisfied by

  • $\star$ $F\subseteq L_\infty (\mu,X)$ and for every $\epsilon>0$ there exists $E\in \Sigma$, $\mu(E)>1-\epsilon$ with $$ \{f\chi_E: f\in F\} $$ is compact in $L_\infty (\mu,X)$.

An example of an $L_1$-compact set that does not satisfy the compactness conditions 2 or $\star$ is the set of monotone step functions $f\colon [0,1]\to \{0,1\}$.

Now for my question: Does there exist a set satisfying 1 and 2 but not satisfying the $L_\infty$ compactness condition $\star$?

I don't have an answer to this for the case $X=R$ and $\Omega=[0,1]$.

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Notice that if $F$ is a set of functions that is compact and metrizable in the topology of pointwise convergence, then for every $\epsilon >0$ there exists $E$, $\mu(E)>1-\epsilon$ such that $$ \\{f\chi_E: f\in F\\} $$ is compact in the topology of uniform convergence. The proof of this is in "Markov Chains" D. Revuz (Proposition 5.5) –  Rabee Tourky Oct 6 '12 at 12:10
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The is (more or less) Egorov's theorem. –  Jochen Wengenroth Oct 9 '12 at 9:11
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@Jochen Well, yes sure it more or less smells like Egorov's theorem. But I've tried and I can't prove that 1 and 2 implies $\star*$. –  Rabee Tourky Oct 9 '12 at 12:47
    
Thanks Jochen. –  Rabee Tourky Oct 15 '12 at 22:52
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1 Answer 1

up vote 3 down vote accepted

The answer is that (1) and (2) implies $\star$ (this resolves a nice problem in a game theory paper I'm working on though the final unresolved problem has to do with decomposable Banach spaces, which I'll ask in a different question).

Assume (1) and (2). Let $P=\{f_n\}$ be a sequence (actual versions of equivalence classes) in $X$ such that each $f\in X$ is the $L_1$ limit of some subsequence of $f_n$. Let $E$ be the measurable set in (2).

By (2), $P$ has compact closure $\overline{P}^p$ in the product topology $X^{E}$. Notice that $X= \overline{P}^{L_1}=\overline{P}^p$ also by (2) taken as equivalence classes. Thus, we can now quickly show by means of Egorov's theorem that for every $\epsilon>0$ there is $F\in \Sigma$, $\mu(F)>1-\epsilon$ such that $\overline{P}^p$ is compact in the topology of uniform convergence (on $F$). We have the result.

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For the record the affirmative answer can be obtained from known techniques: On pointwise convergence, compactness and equicontinuity in the lifting topology. by Ionescu Tulcea (1973) springerlink.com/content/r1452t1107193722 –  Rabee Tourky Oct 18 '12 at 4:25
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