# When can we swap two sigmas in two infinite sums?

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.

• 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.
– Axel
Nov 4, 2022 at 20:18
• @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. Nov 4, 2022 at 20:23
• 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.
– Axel
Nov 4, 2022 at 20:37
• @Axel Apostol Analysis has a reference Nov 4, 2022 at 20:55

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.