0
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ Do you know what the counting measure is? $\endgroup$ – Daniel Fischer Aug 2 '20 at 11:43
  • $\begingroup$ @DanielFischer Isn't it a measure that returns the cardinality of the input set? $\endgroup$ – Xita Meyers Aug 2 '20 at 11:47
  • $\begingroup$ 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). $\endgroup$ – Daniel Fischer Aug 2 '20 at 11:52
  • $\begingroup$ You can use also other general theorems to interchange order of 2 limits. Uniform convergence, existing double limit with some condition and so on. $\endgroup$ – zkutch Aug 2 '20 at 11:55
  • $\begingroup$ @DanielFischer, I know Lebesgue's DCT: 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? $\endgroup$ – Xita Meyers Aug 2 '20 at 11:59
0
$\begingroup$

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 $$\lim\limits_{y \to y_0}\lim\limits_{x \to x_0}f(x,y) = \lim\limits_{(x,y) \to (x_0,y_0)}f(x,y)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.