Fubini for double series: if $\sum_i \sum_j |a_{ij}|$ converges then both iterated sums and the sum along every bijection $\mathbb{N} \to \mathbb{N} \times \mathbb{N}$ converge to one and the same value

Rectangles are bounded by $L$. For all $P, Q \in \mathbb{N}$ one has $\sum_{i<P}\sum_{j<Q}|a_{ij}| \le \sum_{i<P} A_i \le L$, since each inner sum is a partial sum of the convergent nonnegative series $\sum_j |a_{ij}|$ and so is at most $A_i$, and finite sums are monotone.

Single points. Let $d : \mathbb{N}\times\mathbb{N} \to \mathbb{R}$ vanish except at one pair $(p,q)$, let $N \in \mathbb{N}$ and let $\rho$ be injective on $\{n : n<N\}$ with values in $\mathbb{N}\times\mathbb{N}$. If $(p,q) = \rho(n_0)$ for some (necessarily unique) $n_0 < N$, then $\sum_{n<N} d_{\rho(n)} = d_{pq}$; otherwise $\sum_{n<N} d_{\rho(n)} = 0$. Both follow by splitting the sum at $n_0$ and at $n_0+1$, all remaining terms being $0$.

List dominated by a rectangle. For every $N \in \mathbb{N}$, every array $(c_{ij})$ of nonnegative reals, all $P, Q \in \mathbb{N}$ and every injective $\rho$ on $\{n : n<N\}$ with values in $\{(i,j) : i<P,\ j<Q\}$, one has $\sum_{n<N} c_{\rho(n)} \le \sum_{i<P}\sum_{j<Q} c_{ij}$. Induction on $N$, everything else universally quantified: at $N = 0$ the left side is $0$ and the right side is nonnegative; and passing from $N$ to $N+1$, put $(p,q) := \rho(N)$ and let $c''$ agree with $c$ except that $c''_{pq} := 0$, so that the induction hypothesis applied to $c''$ and $\rho$ restricted gives $\sum_{n<N} c_{\rho(n)} \le \sum_{i<P}\sum_{j<Q}c_{ij} - c_{pq}$, the subtraction coming from splitting the outer sum at $p$ and the inner one at $q$; adding $c_{pq}$ closes the induction.

Bounding indices. For every $N$ there are $P, Q$ with $J(n) \in \{(i,j) : i<P,\ j<Q\}$ for all $n < N$; and for all $P, Q$ there is $N$ with $\{(i,j) : i<P,\ j<Q\} \subseteq \{J(n) : n<N\}$. Both are inductions using that the order on $\mathbb{N}$ is total, so that finitely many naturals have a strict upper bound; the second uses surjectivity of $J$ to name, for each pair, the index mapping onto it.

For every $i$ the series $\sum_j a_{ij}$ converges, since $\sum_j |a_{ij}|$ does; write $R_i$ for its sum, so $|R_i| \le A_i$ by [L4] and [L10]. Hence $\sum_i |R_i|$ converges by comparison with $\sum_i A_i$, and $\sum_i R_i$ converges absolutely.

Let $\varepsilon > 0$ be real. Choose $P_0 \ge 1$ with $L - \sum_{i<P_0} A_i < \varepsilon$, possible because the partial sums of $\sum_i A_i$ converge to $L$; then choose, for each $i < P_0$, an index $Q_i$ with $A_i - \sum_{j<Q_i}|a_{ij}| < \varepsilon/P_0$, and let $Q_0$ be an upper bound of the finitely many $Q_i$, so that $A_i - \sum_{j<Q_0}|a_{ij}| < \varepsilon/P_0$ for every $i < P_0$.

Rectangle to list. Let $c$ be an array, let $P, Q, N \in \mathbb{N}$ and let $\rho$ be injective on $\{n : n<N\}$ with $\{(i,j) : i<P,\ j<Q\} \subseteq \{\rho(n) : n<N\}$. Let $c'$ agree with $c$ on that rectangle and vanish off it. Then $\sum_{i<P}\sum_{j<Q} c_{ij} = \sum_{n<N} c'_{\rho(n)}$. This is proved by induction on $P$, with an inner induction on $Q$: enlarging the rectangle by one column adds the single term $c_{PQ}$ to the left side, and changes $c'$ by an array vanishing except at $(P,Q)$, which by step 1.2 adds exactly $c_{PQ}$ to the right side; at $P = 0$ or $Q = 0$ both sides are $0$.

By step 1.3 and step 1.4, every partial sum $\sum_{n<N}|a_{J(n)}|$ is at most $\sum_{i<P}\sum_{j<Q}|a_{ij}| \le L$; hence $\sum_n |a_{J(n)}|$ converges, with sum $\Lambda \le L$, and $\sum_n a_{J(n)}$ converges, say to $S$. Any two bijections $\mathbb{N} \to \mathbb{N}\times\mathbb{N}$ differ by a bijection of $\mathbb{N}$, so by [L9] the value $S$ does not depend on $J$; this is claim 1.

Write $D := \sum_{i<P_0}\sum_{j<Q_0} a_{ij}$ and $E := \sum_{i<P_0}\sum_{j<Q_0} |a_{ij}|$. By step 1.6 and monotonicity, $E > \sum_{i<P_0}(A_i - \varepsilon/P_0) = \sum_{i<P_0}A_i - \varepsilon > L - 2\varepsilon$, so $L - E < 2\varepsilon$.

By step 1.4 fix $N$ with $\{(i,j) : i<P_0,\ j<Q_0\} \subseteq \{J(n) : n<N\}$, and by step 1.4 again fix $P \ge P_0$, $Q \ge Q_0$ with $J(n)$ in the rectangle $\{(i,j) : i<P,\ j<Q\}$ for all $n<N$.

The transposed array $a^{\mathsf{T}}_{ij} := a_{ji}$ satisfies (H): its $i$-th row total is $\sum_j |a_{ji}|$, which converges because its partial sums $\sum_{j<Q}|a_{ji}|$ are bounded by $L$ by step 1.1; and the partial sums $\sum_{i<P}\sum_j |a_{ji}|$ are limits of the rectangle sums $\sum_{i<P}\sum_{j<Q}|a_{ji}|$, again bounded by $L$ by step 1.1, so the series of row totals converges.

For every $N$, $\bigl|S - \sum_{n<N}a_{J(n)}\bigr| \le \Lambda - \sum_{n<N}|a_{J(n)}|$: for $M > N$ the triangle inequality gives $\bigl|\sum_{n<M}a_{J(n)} - \sum_{n<N}a_{J(n)}\bigr| \le \sum_{n<M}|a_{J(n)}| - \sum_{n<N}|a_{J(n)}| \le \Lambda - \sum_{n<N}|a_{J(n)}|$, and letting $M$ grow, the limit preserves the two non-strict inequalities bounding the left side.

Let $a'$ agree with $a$ on the rectangle $\{(i,j) : i<P_0,\ j<Q_0\}$ and vanish off it. By step 2.1, $D = \sum_{n<N} a'_{J(n)}$ and $E = \sum_{n<N} |a'_{J(n)}|$; since $|a'_{J(n)}| \le |a_{J(n)}|$ termwise, monotonicity gives $E \le \sum_{n<N}|a_{J(n)}| \le \Lambda \le L$.

By step 3.1 and step 3.2, $\bigl|S - \sum_{n<N} a_{J(n)}\bigr| \le \Lambda - \sum_{n<N}|a_{J(n)}| \le L - E < 2\varepsilon$.

Also $\bigl|\sum_{n<N}a_{J(n)} - D\bigr| = \bigl|\sum_{n<N}(a - a')_{J(n)}\bigr| \le \sum_{n<N}|(a-a')_{J(n)}| \le \sum_{i<P}\sum_{j<Q}|(a-a')_{ij}| = \sum_{i<P}\sum_{j<Q}|a_{ij}| - E \le L - E < 2\varepsilon$, the middle inequality by step 1.3 and the following equality by splitting the iterated sum at $P_0$ and at $Q_0$, the array $a - a'$ agreeing with $a$ off the small rectangle and vanishing on it.

For each $i < P_0$, $\bigl|R_i - \sum_{j<Q_0}a_{ij}\bigr| \le A_i - \sum_{j<Q_0}|a_{ij}| < \varepsilon/P_0$, by the argument of step 3.1 applied to the row $i$; summing over $i < P_0$ gives $\bigl|\sum_{i<P_0}R_i - D\bigr| < \varepsilon$.

Writing $\Sigma R$ for the sum of $\sum_i R_i$, the same argument applied to the series $\sum_i R_i$ and the comparison $|R_i| \le A_i$ gives $\bigl|\Sigma R - \sum_{i<P_0}R_i\bigr| \le \sum_{i=0}^{\infty}|R_i| - \sum_{i<P_0}|R_i| \le L - \sum_{i<P_0}A_i < \varepsilon$.

Combining step 4.1, step 4.2, step 4.3 and step 4.4, $|\Sigma R - S| < \varepsilon + \varepsilon + 2\varepsilon + 2\varepsilon = 6\varepsilon$. As $\varepsilon > 0$ was arbitrary and $|\Sigma R - S| \ge 0$, this forces $\Sigma R = S$, which with step 1.5 is claim 2.

Applying claims 1 and 2 to $a^{\mathsf{T}}$ and to the bijection $J^{\mathsf{T}}$ obtained by exchanging the coordinates of $J$ gives claim 3, since $a^{\mathsf{T}}_{J^{\mathsf{T}}(n)} = a_{J(n)}$ for every $n$, so the two linear series are the same series and have the same sum $S$.