Basically the point is that it is ambiguous to use an index as both a free and dummy index.
$$ A_iB_{ii} \stackrel{\huge{?}}{=} A_iB_{jj} = A_i \sum_j B_{jj}$$
or
$$ A_iB_{ii} \stackrel{\huge{?}}{=} A_jB_{ji} = \left(\sum_j A_jB_j\right)_i ? $$
This is a problem, we can't have both.
Added after Trapu's edit. So, to be clear, I will be non-einstein notation on the r.h.s. of the equations below: the point here is that $A_iB_{ii}$ cannot be interpreted meaningfully treating one pair of $i$ as dummies (summed over) and the other as free (not summed)
$$ A_iB_{ii} = \sum_{j=1}^n A_jB_{ji} = A_1B_{1i}+A_2B_{2i}+ \cdots +A_nB_{ni} \qquad (I.) $$
verses:
$$ A_iB_{ii} = \sum_{j=1}^n A_iB_{jj} = A_i\sum_{j=1}^n B_{jj} = A_i\left(B_{11}+B_{22}+ \cdots +B_{nn} \right) \qquad (II.) $$
Expressions (I.) and (II.) are two reasonable interpretations of $A_iB_{ii}$ if just one index is taken to be free. But, these are not equal.
For example, $B_{11}=1, B_{22}=-1, B_{12}=0=B_{21}$ and $A_1=1, A_2=1$,
$$ (I.) \qquad A_1B_{1i}+A_2B_{2i} = B_{1i}+B_{2i} = \begin{cases} 1 & i=1 \\ -1 & i=2 \end{cases} $$
verses
$$ (II.) \qquad A_i(B_{11}+B_{22}) = 0 = \begin{cases} 0 & i=1 \\ 0 & i=2 \end{cases} $$
As you can see these expressions do not agree. Therefore, we cannot use the same index for a dummy and a free index.
I would liken this problem to that I have with my calculus I students who insist they need not change the bounds in a u-substitution since they're just going to write it back in terms of $x$ at the end. However, if such a practice is made then some of the intermediate steps are wrong. We are left with the situation that what we write is insufficient to capture the precise mathemtical intent of the expression. This should be avoided since good notation ought to be unambiguous. Or, at a minimum the ambiguity should reflect a deeper mathematical structure as in the case of quotient spaces and the non-uniqueness of the representative. This is not that, this is just bad notation. It does lead to errors, trust me, I've made them.
If $f$ is non-negative, then the fact that
$$\sum_{x\in X} f(x) = \sup \{\sum_{x\in F} f(x); F\text{ is finite}\}$$
tells us that $\int_X f=\sum\limits_{x \in X} f$ by definition of the Lebesgue integral as the $\sup$ of the integral of lower simple functions (not exactly by definition, but every simple function $\sum\limits_{x \in F}s(x)\chi_{\{x\}}$ which is lower than $f$ is lower than $\sum\limits_{x \in F}f(x)\chi_{\{x\}}$, and thus the sup of the definition of the Lebesgue integral in this case is equal to the $\sup$ above).
Thus, if $f$ is integrable, it holds that $\int_X f=\sum\limits_{x \in X} f$.*
If $f$ is not integrable, then wlog we can suppose that $\int_X f^+=+\infty$. This implies that there exists a countable set $C$ inside $X$ on which $f$ is positive and $\sum\limits_{x \in C}f(x)=+\infty$.
Thus, given any $S \in \mathbb{R}$, given $\epsilon=1$ and any finite $F_0 \subset X$ we can take a big enough finite $F'$ (which can be taken disjoint from $F_0$) inside $C$ such that $\sum\limits_{x \in F'}f(x) >S+1-\sum\limits_{x \in F_0} f(x)$, and thus
\begin{align*}
\sum_{x \in F' \cup F_0} f(x)&=\sum_{x \in F'} f(x)+\sum_{x \in F_0} f(x)>S+1,
\end{align*}
contradicting the definition of $\sum\limits_{x \in X} f(x)=S$.
It follows that $f$ is integrable if and only if it is "summable", in the sense you define in your Definition.
*For this to hold, we need to prove that if $\sum\limits_{x \in X} f^+=S_1$ and $\sum\limits_{x \in X} f^-=S_2$, then $\sum\limits_{x \in X} f=S_1-S_2$.
Let $\epsilon>0$. Take a finite set $F_0$ for $f^+$ and $F_1$ for $f^-$, corresponding to $\epsilon/2$ with respect to the definition of summable. Note that we can take those to be disjoint, since they come from the positive and negative parts. Now, let $F \supset F_0 \cup F_1$ be finite. Then
\begin{align*}
|\sum_{x \in F}f-S_1+S_2|&= |\sum_{x \in F-F_1}f-S_1+\sum_{x \in F_1}f+S_2|\\
&< \epsilon,
\end{align*}
as desired.
Alternatively, we can prove that $f \in L^1(X)$ implies that $f$ is summable by reducing to the countable case.
Note that $\operatorname{supp} f :=\{x \in X \mid f(x) \neq 0\}$ is at most countable. This follows by considering $X_n:=\{x \in X \mid |f(x)| \geq 1/n\}$, for then
$$\operatorname{card} X_n \leq \int_{X_n} |nf| = n\int_X |f|<\infty. $$
Since $\operatorname{supp} f=\bigcup X_n,$ it follows that it is countable. Thus, $f$ being integrable allows us to reduce to the countable case and to infer the result that $f \in L^1(X) \implies f$ is summable.
Best Answer
When the definition says "$A$ finite, $A \subset I$", it means exactly what it says: it includes every finite subset of $I$. Really.
You may be troubled because you are trying to reconcile this definition with the prior understanding of $\sum_{n=0}^\infty a_n$ involving the convergence of a sequence of partial sums. But the definition using $\mathrm{sup}$ is not based on the convergence of a sequence partial sums. When we define the set $$ T = \left\{\sum_{i \in A}^{} a_i \mid A \text{ finite}, A \subset I \right\}, $$
we have something involving sums of only some of the $a_i$s (just as partial sums involve only some of the $a_i$s), and each sum is a number, but the numbers are not in sequence. Since the numbers are not in sequence, there is no way that they can converge to anything. That's why we take the ${\mathrm{sup}}$ -- it's not about convergence, it's about putting an upper bound on the numbers (if there is one) and choosing the least possible upper bound out of all possible upper bounds.
You can't "break" the ${\mathrm{sup}}$ of a set by adding numbers to the set that are equal to or less than numbers already in the set. The ${\mathrm{sup}}$ already had to be at least as great as the greater numbers. For example, if $I = \mathbb N,$ then the fact that you are already planning to include $\sum_{i\in\{1,2,3,4,5,6,7\}} a_i$ in your calculations means it doesn't matter whether you consider $\sum_{i\in\{1,7\}} a_i$; since the $a_i$s are all non-negative, $\sum_{i\in\{1,7\}} a_i$ cannot be greater than $\sum_{i\in\{1,2,3,4,5,6,7\}} a_i.$ In short, the fact that the new idea allows us to count sums over subsets of indices with "holes" in the set, and not just sums over subsets without "holes" (subsets of the form "the first $k$ integers"), has no impact on the final answer.