If series is absolutely convergent then $\sum \limits_{n\in I}a_n=\sum \limits_{k=1}^{\infty}\sum \limits_{n\in I_k}a_n.$ Suppose that the series $\sum \limits_{n=1}^{\infty}a_n$ is absolutely convergent and let $I\subseteq \mathbb{N}$ such that $I=\bigsqcup\limits_{k=1}^{\infty}I_k$. Then show that $$\sum \limits_{n\in I}a_n=\sum \limits_{k=1}^{\infty}\sum \limits_{n\in I_k}a_n. \qquad (*)$$
I don't have any idea how to solve it.
I do know that in any absolute convergent series permutation of terms does not change the sum and I guess it should be used somehow in order to prove equality $(*)$. 
Can anyone show the rigorous proof of equality $(*)$, please? 
 A: Suppose for the moment that the result is known to be true for
convergent series of non-negative terms.
If $\sum_{n=1}^\infty a_n$
is an absolutely convergent series of real numbers, define
$a_n = b_n - c_n,$ for all $n \geqslant 1,$ where $c_n = 0$ when
$a_n \geqslant 0$ and $b_n = 0$ when $a_n \leqslant 0.$ Then
$|a_n| = b_n + c_n,$ therefore $\sum_{n=1}^\infty b_n$ and
$\sum_{n=1}^\infty c_n$ are convergent series of non-negative terms,
therefore:
\begin{align*}
\sum_{n \in I}a_n & = \sum_{n \in I}b_n - \sum_{n \in I}c_n \\
& = \sum_{k=1}^\infty\sum_{n \in I_k}b_n -
\sum_{k=1}^\infty\sum_{n \in I_k}c_n \\
& = \sum_{k=1}^\infty\left(
\sum_{n \in I_k}b_n - \sum_{n \in I_k}c_n\right) \\
& = \sum_{k=1}^\infty\sum_{n \in I_k}(b_n - c_n) \\
& = \sum_{k=1}^\infty\sum_{n \in I_k}a_n.
\end{align*}
So it is enough to prove the result on the assumption that
$a_n \geqslant 0$ for all $n \geqslant 1.$
Given any set $K \subseteq \mathbb{N},$ I shall use the Iverson
bracket notation:
$$
[n \in K] =
\begin{cases}
1 & \text{if } n \in K, \\
0 & \text{if } n \notin K.
\end{cases}
$$
I shall assume that, however the notation $\sum_{n \in K}a_n$ has
been defined, it satisfies the identity:
$$
\sum_{n \in K}a_n = \sum_{n=1}^\infty a_n[n \in K].
$$
Let $J_k = I_1 \cup I_2 \cup \cdots \cup I_k$ ($k = 1, 2, \ldots$).
Because the $I_k$ are disjoint, we have
$$
[n \in J_k] =
[n \in I_1] + [n \in I_2] + \cdots + [n \in I_k],
$$
therefore
$$
\sum_{n \in I_1}a_n + \sum_{n \in I_2}a_n + \cdots +
\sum_{n \in I_k}a_n =
\sum_{n \in J_k}a_n \leqslant \sum_{n \in I}a_n,
$$
therefore
$$
\sum_{k=1}^\infty\sum_{n \in I_k}a_n \leqslant \sum_{n \in I}a_n,
$$
and the outer infinite sum on the left hand side exists, because its
partial sums are bounded above by the sum on the right hand side.
On the other hand, for all $m \geqslant 1,$
\begin{align*}
\sum_{n=1}^ma_n[n \in I] & = \sum_{n=1}^ma_n[n \in I_1] +
\sum_{n=1}^ma_n[n \in I_2] + \cdots + \sum_{n=1}^ma_n[n \in I_r] \\
& \leqslant \sum_{n \in I_1}a_n +
\sum_{n \in I_2}a_n + \cdots + \sum_{n \in I_r}a_n \\
& \leqslant \sum_{k=1}^\infty\sum_{n \in I_k}a_n,
\end{align*}
where
$$
r = \max\{k \colon n \leqslant m \text{ for some } n \in I_k\},
$$
therefore
$$
\sum_{n \in I}a_n \leqslant \sum_{k=1}^\infty\sum_{n \in I_k}a_n,
$$
and the two inequalities together prove (*).
A: First assume that $a_n \ge 0$ and define $\sum_{n \in I} a_n = \sup_{J \subset I, J \text{ finite}} \sum_{n \in J} a_n$. Note that it follows that if $I \subset I'$ then
$\sum_{n \in I} a_n  \le \sum_{n \in I'} a_n$.
From https://math.stackexchange.com/a/3680889/27978 we see that if
$K = K_1 \cup \cdots \cup K_m$, a disjoint union, then 
$\sum_{n \in K} a_n = \sum_{n \in K_1} a_n + \cdots + \sum_{n \in K_m} a_n$.
Since $I'=I_1 \cup \cdots \cup I_m \subset I$ we see that
$\sum_{n \in I} a_n \ge \sum_{n \in I'} a_n  = \sum_{k=1}^m \sum_{n \in I_k} a_n$. It follows that
$\sum_{n \in I} a_n \ge \sum_{k=1}^\infty \sum_{n \in I_k} a_n$. This is the
'easy' direction.
Let $\epsilon>0$, then there is some finite $J \subset I$ such that
$\sum_{n\in J} a_n > \sum_{n \in I} a_n -\epsilon$. Since $J$ is finite and the $I_k$ are pairwise disjoint we have $J \subset I'=I_1 \cup \cdots \cup I_m$
for some $m$ and so
$\sum_{k=1}^\infty \sum_{n \in I_k} a_n \ge \sum_{k=1}^m\sum_{n \in I_k} a_n  \ge \sum_{k=1}^m\sum_{n \in J \cap I_k} a_n = \sum_{n\in J} a_n > \sum_{n \in I} a_n -\epsilon$.
(It is not relevant here, but a small proof tweak shows that the result holds true even if the $a_n$ do not have a finite sum.)
Now suppose we have $a_n \in \mathbb{R}$ and $\sum_{n \in I} |a_n|  = \sum_{n=1}^\infty |a_n|$ is finite.
We need to define what we mean by $\sum_{n \in I} a_n$. Note that
$(a_n)_+=\max(0,a_n) \ge 0$ and $(a_n)_-=\max(0,-a_n) \ge 0$. Since
$0 \le (a_n)_+ \le |a_n|$ and $0 \le (a_n)_- \le |a_n|$ we see that
$\sum_{n \in I} (a_n)_+ = \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_+$
and similarly for $(a_n)_-$.
This suggests the
definition (cf. Lebesgue integral) 
$\sum_{n \in I} a_n = \sum_{n \in I} (a_n)_+ - \sum_{n \in I} (a_n)_-$.
With this definition, all that remains to be proved is that
$\sum_{k=1}^\infty \sum_{n \in I_k} a_n = \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_+ - \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_-$ and this follows from
summability and the fact that for each $k$ we have
$\sum_{n \in I_k} a_n = \sum_{n \in I_k} (a_n)_+ - \sum_{n \in I_k} (a_n)_-$.
Note: To elaborate the last sentence, recall that I defined
$\sum_{n \in I_k} a_n$ to be $\sum_{n \in I_k} (a_n)_+ - \sum_{n \in I_k} (a_n)_-$, so all that is happening here is the definition is
applied to $I_k$ rather than $I$. Then to finish, note that if $d_k,b_k,c_k$ are summable and satisfy $d_k=b_k-c_k$ then
$\sum_{k=1}^\infty d_k= \sum_{k=1}^\infty b_k- \sum_{n=1}^\infty c_k$,
where $d_k = \sum_{n \in I_k} a_n$, $b_k = \sum_{n \in I_k} (a_n)_+$ and $c_k = \sum_{n \in I_k} (a_n)_-$.
A: I think there is an elementary proof (one without measure theory), that we can adapt from a similar claim in Apostol's Analysis book. Without loss of generality, $I=\mathbb N$. For each $k\in \mathbb N,\ I_k$ may be regarded as a map from some subset $\{1,2,\cdots,\}\subseteq \mathbb N$, to $\{\sigma_k(1),\sigma_k(2),\cdots,\}$ which may or may not be infinite, so $\sigma_k$ is an injective map from the susbet of $\mathbb N$ of the same cardinality as $|I_k|,$ starting at $1$, to the $\textit{set}\ I_k.$ If $|I_k|=j$, extend $I_k$ to all of $\mathbb N$ by mapping $n\in \mathbb N\setminus \{1,2,\cdots, j\}$ to $\mathbb N\setminus \{\sigma_k(1),\sigma_k(2),\cdots, \sigma_k(j)\}$ injectively and defining $a'_n:=0$ for all $n\in \mathbb N\setminus \{\sigma_k(1),\sigma_k(2),\cdots, \sigma_k(j)\}$. This construction will not affect any of the sums, so without loss of generality, $I_k$ maps $\mathbb N$ to a subset of $\mathbb N$ such that 
$\tag1 I_k\  \text{is injective on}\ \mathbb N $
$\tag2 \text{the range of each}\ I_k \ \text{is a subset of } \ \mathbb N, \text{say}\ P_k $ 
$\tag3 \text{the}\ P_k\ \text{are disjoint}$
Now put $\tag4 b_k(n)=a_{I_{k}(n)}\ \text{and}\ s_k=\sum^\infty_{n=0}b_k(n)$
which is well-defined by $(1)-(3).$ We have to prove that 
$\tag5 \sum^\infty_{k=0}a_k=\sum^\infty_{k=0}s_k$
It's easy to show that the right hand side of this converges absolutely. To find the sum, set $\epsilon>0$ and choose $N$ large enough so that $\sum^\infty_{k=0}|a_k|-\sum^n_{k=0}|a_k|<\frac{\epsilon}{2}$ as soon as $n>N.$ This implies also that 
$\tag6\left|\sum^\infty_{k=0}a_k-\sum^n_{k=0}a_k\right|<\frac{\epsilon}{2}$
Now choose $\{I_1,\cdots, I_r\}$ so that each element of $\{a_1,\dots ,a_N\}$ appears in the sum $\sum^\infty_{n=0}a_{I_{1(n)}}+\cdots +\sum^\infty_{n=0}a_{I_{r(n)}}=s_1+\cdots+ s_r.$ Then, if $n>r,N$ we have 
$\tag 7\left|\sum^n_{k=0}s_k-\sum^n_{k=0}a_k\right|<\sum^\infty_{n=N+1}<\frac{\epsilon}{2}$
Now $(5)$ folllows from $(6)$ and $(7).$
