Which topology for compactness and continuity? I know just enough functional analysis to know that this is probably a simple question, but not enough to actually know the answer.
Let $A$ be the set of nondecreasing functions $f: [a,b] \to [a,b]$.  I know from Helly's selection theorem that 
every sequence $f_n \in A$ has a subsequence that converges almost everywhere to some $f \in A$.
I also have a real valued function $H$ on $A$.  This function is not linear, but I can show that if $f_n \in A$ converges almost everywhere to $f \in A$, then $H(f_n)$ converges to $H(f)$.
I would like a short, simple proof that the function $H$ achieves its maximum on $A$.  The clear route to take is to show that $A$ is compact and $H$ is continuous, and I think I have all the ingredients for this, but I am unclear about which topology to choose to make this all work.  
Thanks!
 A: It seems to me that you don't want to define $H$ on $A$ at all.  Rather, you want to define it on $A/\sim$ where $f\sim g$ if $f=g$ almost everywhere.  Give $A$ the uniform topology and $A/\sim$ the quotient topology.  Then, the two conditions that you state are precisely what you want.  Namely, that $\bar H:A/\sim\to \mathbb{R}$ is continuous because it preserves sequences and $A/\sim$ is compact because every sequence has a convergent subsequence.  
A: A slightly more classical answer than Kieth's answer. 
The set of non-decreasing (monotone) functions (equivalence classes) $M$ of functions from $[0,1]$ to $[0,1]$ is norm-compact in the Banach space $L_1$. (Helly's selection theorem) 
Now take the function $H$ and suppose that $f,g\in A$ differ only on a Lebesgue-null set. Clearly, the constant sequence $f$ converges to the $g$ almost surely. Thus, $H(f) = H(g)$.  
This implies that the function $\tilde H\colon M\to \mathbb{R}$ defined by $\tilde H([f]) =H(f)$, for each equivalence class $[f]\in M$, $f\in A$, is well defined. It is also continuous.  
