Cauchy product $\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$ I have been told that, if $\{a_n\}_{n\in\mathbb{N}}$, $\{a_{-n}\}_{n\in\mathbb{N}^+}$, $\{b_n\}_{n\in\mathbb{N}}$ and $\{b_{-n}\}_{n\in\mathbb{N}^+}$ are absolutely summable complex sequences, then$$\Bigg(\sum_{n=-\infty}^{\infty}a_n\Bigg)\Bigg(\sum_{n=-\infty}^{\infty}b_n\Bigg)=\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$$where $\sum_{n=-\infty}^{\infty}a_n=\sum_{n=0}^{\infty}a_n+\sum_{n=1}^{\infty}a_{-n}$.
I know that if $\{a_n\}_{n\in\mathbb{N}}$ or $\{b_n\}_{n\in\mathbb{N}}$ is absolutely summable, then $(\sum_{n=0}^{\infty}a_n)(\sum_{n=0}^{\infty}b_n)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{n-k}b_k=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}a_{n-k}b_k$, i.e. the proposition is true if $\forall n\leq-1\quad a_n=0=b_n$, and have tried to use that to prove the general case, but I get nothing. I have not studied any theory of measure yet.
I $+\infty$-ly thank you for any help!!!
EDIT: my question had been considered as a duplicate to a question already asked, but that is about the case $(\sum_{n=0}^{\infty}a_n)(\sum_{n=0}^{\infty}b_n)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{n-k}b_k$, which I do know, as I had written. Though, I cannot generalise that to show that $(\sum_{n=-\infty}^{\infty}a_n)(\sum_{n=-\infty}^{\infty}b_n)$, which I hope I correctly write as 
$$(\sum_{k=0}^{\infty}a_k+\sum_{k=1}^{\infty}a_{-k}) (\sum_{k=0}^{\infty}b_k+\sum_{k=1}^{\infty}b_{-k})$$ $$=\sum_{n=0}^{\infty}\sum_{k=0}^{n} a_{n-k}b_k +\sum_{n=0}^{\infty}\sum_{k=0}^{n} a_{n-k}b_{-k-1}+\sum_{n=0}^{\infty}\sum_{k=0}^{n} a_{-n+k-1}b_{k}+\sum_{n=1}^{\infty}\sum_{k=1}^{n} a_{-n+k-1}b_{-k},$$
is equal to $\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$. The people who regarded this question of mine as a duplicate to that were enough to have this question closed for a while, therefore, I suppose that it may well be trivial that the equality $(\sum_{n=-\infty}^{\infty}a_n)(\sum_{n=-\infty}^{\infty}b_n)=\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$ derives from the equality, known by me, $(\sum_{n=0}^{\infty}a_n)(\sum_{n=0}^{\infty}b_n)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{n-k}b_k$. If that is trivial, I do not realise that fact: could anybody, either from those considering the question identical to that or anybody else, please, show me how to derive the equality with the indices $n$ from $-\infty$ to $+\infty$? I... $+\infty$-ly thank you! ;-)
 A: This is about $L^1({\mathbb Z}^2,\#)$.
When $S$  is a countable set ($S={\mathbb Z}^2$ in our example) a function $f:\ S\to{\mathbb C}$ is summable when 
$$\sup_{J\subset S,\ J\ {\rm finite}}\sum_{x\in J} |f(x)|<\infty\ .$$
If this is the case then it follows from the completeness of ${\mathbb C}$ that there is a unique $s\in{\mathbb C}$ with the following property: For each $\epsilon>0$ there is a finite set $J_0\subset S$ such that for all finite sets $J$ with $J_0\subset J\subset S$ one has
$$\left|\sum_{x\in J}f(x)-s\right|<\epsilon\ .$$
Denote this number $s$ by $\int_Sf$.
Given two absolutely convergent complex series $$\sum_{k\in{\mathbb Z}}a_k=:A, \quad \sum_{k\in{\mathbb Z}}b_k=:B$$ 
it is easy to check that
$$f:\ {\mathbb Z}^2\to{\mathbb C},\qquad(j,k)\mapsto a_j\>b_k$$
is summable and that
$$\int_{{\mathbb Z}^2}f=A\cdot B\ .$$
On the other hand, for summable $f$ we have a Fubini theorem: Assume that $f$ is summable over $S$, and that $S=\bigcup_{\iota\in I} S_\iota$ is any (maybe infinite) partition of $S$. Then 
$$\int_S f=\sum_{\iota\in I} s(\iota)\ ,$$
where $$s(\iota)=\int_{S_\iota} f\ .$$
Now apply this to the partition
$${\mathbb Z}^2=\bigcup_{n=-\infty}^\infty\bigl\{(n-k,k)\>|\> k\in{\mathbb Z}\bigr\}\ .$$
A: Because the series are absolutely convergent, then so are $a(z)=\sum_{n=-\infty}^{\infty}a_{n}z^{n}$ and $b(z)=\sum_{n=-\infty}^{\infty}b_{n}z^{n}$ absolutely convergent for $|z|=1$. Therefore, the following is also absolutely convergent for $|z|=1$:
$$
          a(z)b(z) = \sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}a_{n}z^{n}b_{m}z^{m}.
$$
So it is permissible to arbitrarily rearrange the terms of the series, without changing the value of the sum. In particular, one may collect all like powers of $z$ without changing the value of the sum. The coefficient of $z^{k}$ is $\sum_{n=-\infty}^{\infty}a_{n}b_{k-n}$ because $n+(k-n)=k$. Therefore,
$$
            a(z)b(z) = \sum_{k=-\infty}^{\infty}\left[\sum_{n=-\infty}^{\infty}a_{n}b_{k-n}\right]z^{k}.
$$
The powers of $z$ were introduced only to illustrate how to collect the terms, and to know that none of the terms were omitted from the final sum. Finally, set $z=1$.
