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Let $f: \mathbb{N} \times \mathbb{N} \to \overline{\mathbb{R}}$. Then under which conditions is the expression $\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty f(m,n)=\sum\limits_{m=1}^\infty \lim\limits_{n\to\infty} f(m,n)$ valid?

Would anyone have a rigorous answer to this? Any proof using measure theory, or elementary calculus, is more than welcome.

I know that a very similar question has been asked here: Under what condition we can interchange order of a limit and a summation? , but I would need more detail. For example, one of the answers states that the dominated convergence theorem suffices as 'sums are just integrals with respect to the counting measure on $\mathbb{N}$'. I am unable to see how works; I don't know how this 'counting measure' can be used with the dominated convergence theorem to provide the conclusion.

Arctic Char
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Xita Meyers
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  • Do you know what the counting measure is? – Daniel Fischer Aug 02 '20 at 11:43
  • @DanielFischer Isn't it a measure that returns the cardinality of the input set? – Xita Meyers Aug 02 '20 at 11:47
  • Yes (except, to be pedantic, for all infinite sets, whatever their cardinality, the value is $+\infty$, not a cardinal number). Okay, so that's not the point of confusion. Did you learn the dominated convergence theorem only for the Lebesgue measure? It holds for all measures, hence also for the counting measure (on whatever set). – Daniel Fischer Aug 02 '20 at 11:52
  • You can use also other general theorems to interchange order of 2 limits. Uniform convergence, existing double limit with some condition and so on. – zkutch Aug 02 '20 at 11:55
  • @DanielFischer, I know Lebesgue's DCT: https://en.wikipedia.org/wiki/Dominated_convergence_theorem. The point of confusion is, I don't know how 'sums are just integrals with respect to the counting measure on $\mathbb{N}$'. No matter how obvious, could I ask you to spell things out a bit more? – Xita Meyers Aug 02 '20 at 11:59
  • @zkutch Would you be suggesting I use Fubini-Tonelli theorem? – Xita Meyers Aug 02 '20 at 12:09
  • A sequence is just a function $\mathbb{N} \to \mathbb{R}$ (or some other space for the codomain). If you look at the counting measure $\nu$ on $\mathbb{N}$, a simple function is just a sequence with only finitely many non-zero terms. And every singleton set has measure $1$. Thus the integral of a simple function $f$ with respect to $\nu$ is $$ \int f,d\nu = \sum_{n \in \mathbb{N}} f(n)\nu({n}) = \sum_{n \in \mathbb{N}} f(n),.$$ And if you extend that to arbitrary $\nu$-integrable functions (note that the relevant $\sigma$-algebra is the power set of $\mathbb{N}$, so measurability isn't – Daniel Fischer Aug 02 '20 at 12:12
  • an issue) just like for any other measure, you get the same formula $$\int f,d\nu = \sum_{n \in \mathbb{N}} f(n),.$$ Note that $\nu$-integrability is the same as absolute convergence, so this viewpoint doesn't help if you're dealing with series that are only conditionally convergent. – Daniel Fischer Aug 02 '20 at 12:12
  • Fubini mainly I know about integrals, but here you have simple double sequence. There are several conditions which give order interchanging. – zkutch Aug 02 '20 at 12:14

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While looking for higher considerations let me suggest some simple criteria.

  1. Suppose we have double sequense $a_{n,m}$. If exists $\lim_\limits{m \to \infty}a_{n,m}=a_n, \ n\in \mathbb{N}$ ; and series $\sum\limits_{n=1}^{\infty}a_{n,m}$ converged uniformly, then we can interchange order of limit and summation.

  2. Suppose series $\sum\limits_{m,n =1}^{\infty}a_{n,m}$, $\sum\limits_{n =1}^{\infty}\sum\limits_{m =1}^{\infty}a_{n,m}$ and $\sum\limits_{m =1}^{\infty}\sum\limits_{n =1}^{\infty}a_{n,m}$ all converged. Then they equal one and same value.

  3. Suppose $f(x,y)$ is defined on some set $E$, which includes all points from some rectangle with center in $(x_0,y_0)$, except, possibly, lines $y=y_0$ and $x=x_0$. If exists double limit for $f$ with respect to $E$ and for any $y \ne y_0$ in some neigbourhood of $y_0$ exists $\lim\limits_{x \to x_0}f(x,y) = g(y)$, then exists $\lim\limits_{y \to y_0}g(y)$ and holds $$\lim\limits_{y \to y_0}\lim\limits_{x \to x_0}f(x,y) = \lim\limits_{(x,y) \to (x_0,y_0)}f(x,y)$$

zkutch
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