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 $\Omega=[0,1]$.



Certain compact subset of $L_1$
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. 


