Exchange of Limit and Integral with Nets

Question

In topology, we have seen that there are examples of nets so that monotone and dominated convergence do not hold anymore.

In particular, we worked with the net $\mathfrak{F}$ containing finite subsets of $[0,1]$ ordered by inclusion. We used the Lebesgue-measure $\lambda$ restricted to $[0,1]$.

The net $(\chi_F)_{F \in \mathfrak{F}}$ is monotonically increasing and is dominated by $\chi_{[0,1]}$. $(\chi_F)_{F \in \mathfrak{F}}$ converges pointwise to $\chi_{[0,1]}$ as well.

But $\lim_{F \in \mathfrak{F}} \int \chi_F d\lambda ≠ \int \chi_{[0,1]} d\lambda$.

Are there properties/constraints of the net itself (except the ones elaborated by David C. Ullrich below) or of the measure that expand the exchangeability of limit and integral to nets?

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Some of my thoughts on monotone convergence: The problem is that we cannot sort the functions we have in order to make it monotonous as one is used to when dealing with natural numbers as an index. But we do have a directed set at least. Do you have any ideas how to constrain the functions in the net in order to get a behavior that is similar to monotonicity? I have the impression it is, because monotonous functions only have countable points of discontinuity calling them $(a_n)_{n \in \mathbb{N}}$, at least in $\mathbb{R}$, which might be used for collecting some functions of the net.

Defining a set: $\{]a_i, \infty[;$$a_i$ point of discontinouity$\}$ and taking a look at its $\sigma$-Algebra, introducing an ordering there via inverse inclusion ($ A ≤ B \Leftrightarrow A \supseteq B$). Adding ${\infty}$ to this set, we might find a directed set. (just some thoughts... to be continued)

Answer 1

I doubt that there's a natural condition on uncountable nets that's going to work. Countable nets are ok; whether the argument below shows this is "obvious" will depend on the reader.

Countable nets are ok because you really don't get anything out of countable nets that you can't get from using sequences.

More precisely, any countable directed set contains an increasing cofinal sequence.

Expanding on that: Say $A=(\alpha)$ is a countable directed set. There is a sequence $(a_n)\subset A$ such that $\alpha_{n+1}\ge \alpha_n$ and such that for every $\alpha\in A$ there exists $n$ with $\alpha_n\ge \alpha$.

(Hence $\lim_\alpha x_\alpha=L$ implies $\lim_nx_{\alpha_n}=L$.)

Proof. Say $A=(\beta_1,\beta_2\,\dots)$. Let $\alpha_1=\beta_1$. Having chosen $\alpha_n$, choose $\alpha_{n+1}\in A$ such that $\alpha_{n+1}\ge\alpha_n$ and also $\alpha_{n+1}\ge\beta_k$, $1\le k\le n$.