In the proof of the Kreps-Yan theorem it is pointed out that the following set is weakly compact (for the dual pair $<L^1,L^\infty>$)
$B_\delta:=\{ f\in L^1 \:;\: 0\leq f\leq 1,\: E[f]\geq\delta \}$,
where $L^1=L^1(\Omega,\mathcal{F},P)$ for some probability space $(\Omega,\mathcal{F},P)$.
How could easily prove that $B_\delta$ is weakly compact?
A posibility is to use the Kames' compactness theorem.
First, $B_\delta$ is bounded, convex and closed (in $L^1$).
Second, for each $z\in L^\infty$, we have to see that $y\mapsto E[y z]$ attains a maximum in $B_\delta$. Indeed, let $A:=(z\geq 0)$.
If $P(A)\geq \delta$, then it is easy to see that $E[1_A z]=\underset{y\in B_\delta}\max E[y z]$.
If $P(A)<\delta$, we see that $1_A\notin B_\delta$, since $E[1_A]<\delta$. Then, we put $\alpha:=\frac{\delta - P(A)}{P(A^c)}$, and we have
$E[(1_A + \alpha 1_{A^c}) z]=\underset{y\in B_\delta}\max E[y z]$.