compactness of distribution functions I came up with the following assertion and I am  having hard time to justify. The author says the following: 
"Consider a closed and convex set of probability measures on a compact set, say $[0,1]$. Denote this set by $\Delta$. Assume that each measure in this set has a strictly increasing and continuous cumulative distribution function. Denote this set by $\Delta_{CDF}$. Then, since $\Delta$ is compact under weak-* topology, the following function is well-defined:
$
F_{\min }\left( x\right) =\min_{F\in \Delta _{CDF}}F\left( x\right)
$
"
First of all, how can one assume "each measure on $\Delta$ has a strictly increasing and continuous cumulative distribution function"? Is this something legitimate? That is, can there be a closed and convex set of measures such that every measure has a s.increasing and continuous cdf?
Second, does this mean the answer to the following question is "yes"? If so, what topology?
Consider the set of strictly increasing and continuous cumulative distribution functions on a compact set, say $[0,1]$. Is there a topology under which this set is compact?
 A: 1)Consider any two continuous strictly increasing CDF's $\{F_1, F_2\}$ of measures on $[0,1]$. 
For instance consider $F_1(x) = x$ and $F_2 (x) = x^2$ ($x \in[0,1]$)
Therefore any convex combination of the corresponding measures (say $\mu_1, \mu_2$) is a measure with continuous strictly increasing CDF (indeed $\mu_\alpha = \alpha \mu_1 + (1-\alpha) \mu_2 \Rightarrow F_\alpha (x) = \mu_\alpha ([0,x]) = \alpha F_1(x) + (1-\alpha) F_2(x)$ is a continuous function)
Consider the set $[F_1,F_2] = \{\mu_\alpha, \alpha \in[0,1]\}$ this set is convex and every measure in it has a continuous strictly increasing CDF.
2)The question now is :
Consider the set of strictly increasing and continuous cumulative distribution functions on a compact set, say [0,1]. Is there a topology under which this set is compact?
The answer is yes, just take the trivial topology $\mathcal{O}_T$ the set $\Delta_{CDF}$ is compact. 


Consider the topology of weak convergence (induced by the Lévy-Prokhorov metric $$d(\mu, \nu) = \inf\{\epsilon>0:F_\mu (x) \leq F\nu(x+ \epsilon) + \epsilon , F_\nu (x) \leq F\mu(x+ \epsilon) + \epsilon \}$$
 where $F_\mu(x) = \mu([0,x])$ is the cumulative distribution of $\mu$)
Under this topology, every sequence of measures with support on the compact set $[0,1]$  admits a convergent subsequence (cf tightness). But if you consider the family  of piecewise linear functions $\{F_n\}_n$ 
$$F_n(x) = \begin{cases}0 &\text{if } x  =  0 \\
1/n& \text{if } x  =  1/2\\
1 - 1/n& \text{if } x  =  1/2 + 1/n\\
1 & \text{if } x  =  1\\
\end{cases}$$
One can see that each $F_n$ is continuous and strictly increasing CDF's.
The $F_n$ converge in distribution to $\delta_{1/2}$ wich is not a measure in $\Delta_{CDF}$. Therefore convergence in the topology we are going to consider on this set does not imply convergence in distribution. 
Since $\{F_n\}$ is in $\Delta_{CDF}$ and in the topology we are going to converge to some $F$ where could it be converging to?


Well consider the trivial topology $\mathcal{O}_T = \{\emptyset, \Delta_{CDF}\}$. In this  topology  the set of $\Delta_{CDF}$ is compact. since the every open covering  of $\Delta_{CDF}$ admits a finite subcover.
This topology  might not be interesting. I suspect that for any interesting topology the above sequence $\{F_n\}$ converges to $\delta_{1/2}$ and so there would be no interesting topology for wich $\Delta_{CDF}$ is compact. 
Anyway, I hope this sheds some light in your problem.
