# Exchange of Limit and Integral with Nets [duplicate]

This question already has an answer here:

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?

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)

## marked as duplicate by Alex M., Claude Leibovici, Hans Engler, Ethan Bolker, StrantsMay 23 '18 at 14:31

• +1 for pointing out that DCT and MCT fail for nets! I know of at least one excellent book on real analysis (Folland) that includes a section on nets (for the Tychonoff theorem, mainly) but doesn't mention this fact. It should. Regarding your question, I don't know. Countable nets are ok, but it's not clear to me whether that's "obvious"; it's very easy but not entirely trivial... – David C. Ullrich Jul 17 '15 at 18:48
• @DavidC.Ullrich The very obvious for me were just sequences with properties so that LDCT or LMCT holds (which depends on the measure too). – mdot Jul 17 '15 at 18:52
• Hmm. Not clear to me whether that means you were aware of what I said in the answer I just posted or not... – David C. Ullrich Jul 17 '15 at 19:04

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 $\alpha_{n+1}\ge\beta_k$, $1\le k\le n$.

• I made the mistake of applying LDCT to a problem with nets (regarding a lemma related to Bochner's theorem) in a meeting with an advisor. I was incredibly embarrassed that I made such an egregious error. I think countable nets is probably the best you can do in general. – Cameron Williams Jul 17 '15 at 19:06
• I never made the same mistake in front of anyone, but when I learned about nets I just assumed that MCT and DCT were ok. Until I thought about it somewhat later... – David C. Ullrich Jul 17 '15 at 19:09