We know the following monotone convergence theorem:
If $n:X\to\mathbb R$ is a bounded increasing net, then $n$ converges. (Also for other spaces than $\mathbb R$.)
I am trying to formulate this using filters, i.e., something like "if $F$ is a bounded increasing filter, then $F$ converges".
It is often said, that all theorems about nets can be stated in terms of filters and vice-versa, but I struggle figuring out how to even formulate the monotone convergence theorem using filters. (In particular, I do not know how an "increasing filter" should be defined.)
How to formulate the monotone convergence theorem using filters?
(In case it matters: I want a formulation that is not specific to $\mathbb R$ because I also want to formulate it also for bounded operators.)
Here is one reasonable way to generalize this to filters. If $X$ is a partially ordered set and $F$ is a filter on $X$, say that $F$ is increasing if $\{x\in X:[x,\infty)\in F\}\in F$. In other words, for almost every $x$ (where "almost every" is interpreted using $F$), $[x,\infty)\in F$.
If $F$ is the filter on $X$ induced by some net $(x_i)_{i\in I}$, then $F$ is increasing in this sense iff $(x_i)$ is "weakly increasing" in the sense that there exists $i_0\in I$ such that for each $i\geq i_0$ there exists $j\in I$ such that $x_k\geq x_i$ for all $k\geq j$. Indeed, $[x_i,\infty)\in F$ iff such a $j$ exists for $i$ so this condition says exactly that the net is eventually in the set $\{x\in X:[x,\infty)\in F\}$.
Also, a filter $F$ is increasing iff it contains the filter induced by some net that is increasing in the usual sense. One direction is trivial since any refinement of an increasing filter is still increasing. In the other direction, suppose $F$ is increasing and let $I=\{x\in X:[x,\infty)\in F\}$. Note that $I$ is directed (since if $x,y\in I$ then $[x,\infty)\cap[y,\infty)\in F$ so it must intersect $I$) so we can consider the identity net $I\to X$. This is an increasing net and $\{j\in I:j\geq i\}$ is in $F$ for any $i\in I$ since this set is equal to $I\cap[i,\infty)$. So, $F$ contains the filter induced by this net.
Now we can prove the following theorem.
Theorem: Suppose $X$ is a complete totally ordered set with a greatest element and $F$ is an increasing filter on $X$. Then $F$ converges in the order topology of $X$.
(The condition that $X$ has a greatest element is just the analogue of the requirement that the net be bounded above. If $X$ doesn't have a greatest element, you can instead just require that some element of $F$ is bounded above, so you can replace $X$ with $(-\infty,a]$ for some $a\in X$ and then apply the statement above to the filter restricted to $(-\infty,a]$.)
Proof: Let $S=\{x\in X:[x,\infty)\in F\}$ and let $a=\sup S$ (which exists since $X$ is complete and has a greatest element). I claim that $F$ converges to $a$. To show this, we must show that $(x,\infty)\in F$ for all $x<a$ and $(-\infty,x)\in F$ for all $x>a$. First, if $x<a$, then there exists $s\in S$ such that $x<s\leq a$, so $(x,\infty)\supseteq[s,\infty)\in F$. On the other hand, if $x>a$, then $(-\infty,x)\supseteq (-\infty,a]\supseteq S\in F$ since $F$ is increasing. Thus $F$ converges to $a$.