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I have a space of elementary outcomes $\Omega = \{w_1, w_2, w_3, ... \}$

We have assigned a non-negative number $p(w)$ to every elementary outcome $w \in \Omega$ in such a way that

$\sum_{k=1}^\infty p(w_k) = 1$

Can I always swap sigmas in double summations, when both sums are infinite? What about when both sums are finite? Or when we have a finite and an infinite sum there?

In particular, I am mostly interested if in this sum I can swap the two sigmas:

$\sum_{n=1}^\infty \sum_{k=1}^\infty p(w_k) [w_k \in B_n] \tag{*}$

Here (if it matters) $B_n$ are mutually (pairwise) disjoint events (subsets of $\Omega$), and $p(w_k)$ are these non-negative numbers which are assigned to the elementary outcomes (the elements of $\Omega$).

Also, the Iverson bracket is used there.

Can I swap (exchange) the two sigmas in the sum $(*)$? If so, why? In what cases would I not be allowed to swap the two sigmas?

Note: Sorry for the multiple questions. I just want to understand this matter in details.

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    $\begingroup$ You can take a look at en.wikipedia.org/wiki/Fubini%27s_theorem ; you need absolute convergence of $\sum_{n,m} a_{n,m}$. In your case as everything is non-negative, when $\sum_n \sum_k a_{n,k}$ converges you can swap the sums as you like. $\endgroup$
    – Axel
    Nov 4, 2022 at 20:18
  • $\begingroup$ @Axel Thanks. Yes, all the elements of the double sum are non-negative, and also I know that the inner sums themselves are convergent series. But is this enough to satisfy the condition of Fubini's theorem? I am asking because that conditions talks about Cartesian product there. $\endgroup$ Nov 4, 2022 at 20:23
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    $\begingroup$ You're welcome, it suffices that either $\sum_n \sum_k a_{n,k}$ converges or $\sum_k \sum_n a_{n,k}$ converges to swap the sums when $(a_{n,k})$ is a doubly-indexed non-negative sequence. I cannot find a good reference in English for the moment though. $\endgroup$
    – Axel
    Nov 4, 2022 at 20:37
  • $\begingroup$ @Axel Apostol Analysis has a reference $\endgroup$
    – Andrew
    Nov 4, 2022 at 20:55

1 Answer 1

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Let me try to answer your questions :


  • Can I always swap sigmas in double summations, when both sums are infinite?

You cannot always do this, for example you can take the sequence defined by, $$\forall (p,q) \in \Bbb{N}^2, \; a_{p,q} := \delta_{p,q}-\delta_{p+1,q} $$ where $\delta_{i,j} = 1$ when $i =j$ and $0$ otherwise.

If $p \geq 1$ is fixed then $\sum_{q=1}^{\infty}a_{p,q} = a_{p,p}+a_{p,p+1}=1-1=0$

If $q \geq 1$ is fixed :

  • if $q \geq 2, \sum_{p=1}^{\infty} a_{p,q} = a_{q,q}+a_{q-1,q} = 1-1=0$
  • if $q = 1, \sum_{p=1}^{\infty} a_{p,q} = a_{q,q} = 1 $

Therefore,

$$\sum_{p = 1}^{\infty} \sum_{q=1}^{\infty} a_{p,q} = 0 $$

$$\sum_{q = 1}^{\infty} \sum_{p=1}^{\infty} a_{p,q} = 1$$

Hence we cannot swap the sums.

However, when the sequence is non-negative you can always swap the sums. It is also the case when, $$ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} |a_{n,m}|<+\infty \quad \text{ or } \quad \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} |a_{n,m}|<+\infty $$ i.e. when $(a_{n,m})$ is summable, cf. down below.


  • What about when both sums are finite?

You can rearrange them as you want, there is no problem.


  • Or when we have a finite and an infinite sum there?

Well in that case, if you know that,

$$ \sum_{n = 1}^{\infty} a_n \text{ exists }$$ then it is the exact same thing as manipulating limits : $$ \sum_{i = 1}^N b_i \sum_{n = 1}^{\infty} a_n = \sum_{n= 1}^{\infty} \sum_{i = 1}^N b_i a_n $$

However the converse is not true.


  • Can I swap (exchange) the two sigmas in the sum (∗)? If so, why? In what cases would I not be allowed to swap the two sigmas?

In your case you can perfectly swap the sigmas, because the sequence is non-negative. The best I could do was to give you some insight on what I know. All the following results are from a course I had, named summable families. ($I$ will denote a countable set).

Definition

Let $I$ be a countable set and $(a_i)_{i \in I}$ be a family of non-negative real numbers. The family $(a_i)_{i \in I}$ is said to be summable when, $$M := \left\{ \sum_{i \in J} a_i : J \text{ finite subset of } I\right\} $$ has an upper bound, in which case we call sum of $(a_i)_{i \in I}$ the real number : $$\sum_{i \in I} a_i := \sup M$$

The important theorem of this course is the following :

Theorem

Let $(I_{\lambda})_{\lambda \in \Lambda}$ be an at most countable family of subsets of $I$ such that, $$ I = \bigsqcup_{\lambda \in \Lambda} I_{\lambda}$$ A family $(a_i)_{i \in I}$ of non-negative real numbers is summable if, and only if, for all $\lambda \in \Lambda$, $(a_i)_{i \in I_{\lambda}}$ is summable with sum $S_{\lambda}$ and the family $(S_{\lambda})_{\lambda \in \Lambda}$ is summable. When these conditions are verified, the following equality holds, $$\sum_{i \in I} a_i=\sum_{\lambda \in \Lambda} S_{\lambda} = \sum_{\lambda \in \Lambda}\left( \sum_{i \in I_{\lambda}} a_i \right) $$

The idea is that you can sum by grouping the terms. However, the proof is rather tedious to write. As a corollary, it yields Fubini's theorem for infinite series in the non-negative case.

Theorem

A doubly-indexed non-negative sequence $(a_{n,m})_{(n,m) \in \Bbb{N}^2}$ is summable if and only if, one of the following equivalent assertions holds :

  1. for all $n \geq 0$, $\sum_m a_{n,m}$ is convergent and $\displaystyle \sum_n \sum_{m =1}^{\infty} a_{n,m}$ is also convergent ;
  2. for all $m \geq 0$, $\sum_n a_{n,m}$ is convergent and $\displaystyle \sum_m \sum_{n =1}^{\infty} a_{n,m}$ is also convergent.

In which case, $$\sum_{(n,m) \in \Bbb{N}^2} a_{n,m} = \sum_{n=1}^{\infty} \left(\sum_{m = 1}^{\infty} a_{n,m}\right) = \sum_{m=1}^{\infty} \left(\sum_{n = 1}^{\infty} a_{n,m}\right) $$

Proof :

We apply the previous theorem with $I = \Bbb{N}^2$ and the respective partitions of $I$,

  • $(I_n^{(1)})_{n \in \Bbb{N}}$ with $I_n^{(1)} = \{(n,m) : m \in \Bbb{N}\} $
  • $(I_m^{(2)})_{m \in \Bbb{N}}$ with $I_m^{(2)} = \{(n,m) : n \in \Bbb{N}\} $

If you want a more direct proof you can check this post : Zeta question - prime zeta. Basic calculus

More generally : a complex-valued family $(a_i)_{i \in I}$ is said to be summable when $(|a_i|)_{i \in I}$ is summable. And in the case of summability, Fubini's theorem holds, namely $$\sum_{(n,m) \in \Bbb{N}^2} a_{n,m} = \sum_{n=1}^{\infty} \left(\sum_{m = 1}^{\infty} a_{n,m}\right) = \sum_{m=1}^{\infty} \left(\sum_{n = 1}^{\infty} a_{n,m}\right)$$ but in the complex case, there is no such equivalence as in the non-negative one.


I hope this answer can bring some clarity on swapping the sum symbols.

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