Fubini for double series: if ${\sum }_i{\sum }_j\mathrm{\mid }{a}_{ij}\mathrm{\mid }$ 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 ${\sum }_{i<P}{\sum }_{j<Q}\mathrm{\mid }{a}_{ij}\mathrm{\mid }\le {\sum }_{i<P}{A}_i\le L$, since each inner sum is a partial sum of the convergent nonnegative series ${\sum }_j\mathrm{\mid }{a}_{ij}\mathrm{\mid }$ 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 $c^{\mathrm{\prime }\mathrm{\prime }}$ agree with $c$ except that $c_{pq}^{\mathrm{\prime }\mathrm{\prime }}:=0$, so that the induction hypothesis applied to $c^{\mathrm{\prime }\mathrm{\prime }}$ 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 ${\sum }_j\mathrm{\mid }{a}_{ij}\mathrm{\mid }$ does; write $R_i$ for its sum, so $\mathrm{\mid }{R}_i\mathrm{\mid }\le A_i$ by [L4] and [L10]. Hence ${\sum }_i\mathrm{\mid }{R}_i\mathrm{\mid }$ converges by comparison with ${\sum }_i{A}_i$, and ${\sum }_i{R}_i$ converges absolutely.

Let $\epsilon >0$ be real. Choose $P_0\ge 1$ with $L-{\sum }_{i<P_0}{A}_i<\epsilon$, 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 $A_i-{\sum }_{j<Q_i}\mathrm{\mid }{a}_{ij}\mathrm{\mid }<\epsilon \mathrm{/}{P}_0$, and let $Q_0$ be an upper bound of the finitely many $Q_i$, so that $A_i-{\sum }_{j<Q_0}\mathrm{\mid }{a}_{ij}\mathrm{\mid }<\epsilon \mathrm{/}{P}_0$ 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 $c^{\mathrm{\prime }}$ agree with $c$ on that rectangle and vanish off it. Then ${\sum }_{i<P}{\sum }_{j<Q}{c}_{ij}={\sum }_{n<N}{c}_{\rho (n)}^{\mathrm{\prime }}$. 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 $c^{\mathrm{\prime }}$ 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 ${\sum }_{n<N}\mathrm{\mid }{a}_{J(n)}\mathrm{\mid }$ is at most ${\sum }_{i<P}{\sum }_{j<Q}\mathrm{\mid }{a}_{ij}\mathrm{\mid }\le L$; hence ${\sum }_n\mathrm{\mid }{a}_{J(n)}\mathrm{\mid }$ converges, with sum $\mathrm{\Lambda }\le L$, 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 $E:={\sum }_{i<P_0}{\sum }_{j<Q_0}\mathrm{\mid }{a}_{ij}\mathrm{\mid }$. By step 1.6 and monotonicity, $E>{\sum }_{i<P_0}(A_i-\epsilon \mathrm{/}{P}_0)={\sum }_{i<P_0}{A}_i-\epsilon >L-2\epsilon$, so $L-E<2\epsilon$.

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

For every $N$, $\mid S-{\sum }_{n<N}{a}_{J(n)}\mid \le \mathrm{\Lambda }-{\sum }_{n<N}\mathrm{\mid }{a}_{J(n)}\mathrm{\mid }$: for $M>N$ the triangle inequality gives $\mid {\sum }_{n<M}{a}_{J(n)}-{\sum }_{n<N}{a}_{J(n)}\mid \le {\sum }_{n<M}\mathrm{\mid }{a}_{J(n)}\mathrm{\mid }-{\sum }_{n<N}\mathrm{\mid }{a}_{J(n)}\mathrm{\mid }\le \mathrm{\Lambda }-{\sum }_{n<N}\mathrm{\mid }{a}_{J(n)}\mathrm{\mid }$, and letting $M$ grow, the limit preserves the two non-strict inequalities bounding the left side.

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

By step 3.1 and step 3.2, $\mid S-{\sum }_{n<N}{a}_{J(n)}\mid \le \mathrm{\Lambda }-{\sum }_{n<N}\mathrm{\mid }{a}_{J(n)}\mathrm{\mid }\le L-E<2\epsilon$.

Also $\mid {\sum }_{n<N}{a}_{J(n)}-D\mid =\mid {\sum }_{n<N}(a-a^{\mathrm{\prime }}{)}_{J(n)}\mid \le {\sum }_{n<N}\mathrm{\mid }(a-a^{\mathrm{\prime }}{)}_{J(n)}\mathrm{\mid }\le {\sum }_{i<P}{\sum }_{j<Q}\mathrm{\mid }(a-a^{\mathrm{\prime }}{)}_{ij}\mathrm{\mid }={\sum }_{i<P}{\sum }_{j<Q}\mathrm{\mid }{a}_{ij}\mathrm{\mid }-E\le L-E<2\epsilon$, 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 $a-a^{\mathrm{\prime }}$ agreeing with $a$ off the small rectangle and vanishing on it.

For each $i<P_0$, $\mid R_i-{\sum }_{j<Q_0}{a}_{ij}\mid \le A_i-{\sum }_{j<Q_0}\mathrm{\mid }{a}_{ij}\mathrm{\mid }<\epsilon \mathrm{/}{P}_0$, by the argument of step 3.1 applied to the row $i$; summing over $i<P_0$ gives $\mid {\sum }_{i<P_0}{R}_i-D\mid <\epsilon$.

Writing $\mathrm{\Sigma }R$ for the sum of ${\sum }_i{R}_i$, the same argument applied to the series ${\sum }_i{R}_i$ and the comparison $\mathrm{\mid }{R}_i\mathrm{\mid }\le A_i$ gives $\mid \mathrm{\Sigma }R-{\sum }_{i<P_0}{R}_i\mid \le {\sum }_{i=0}^{\mathrm{\infty }}\mathrm{\mid }{R}_i\mathrm{\mid }-{\sum }_{i<P_0}\mathrm{\mid }{R}_i\mathrm{\mid }\le L-{\sum }_{i<P_0}{A}_i<\epsilon$.

Combining step 4.1, step 4.2, step 4.3 and step 4.4, $\mathrm{\mid }\mathrm{\Sigma }R-S\mathrm{\mid }<\epsilon +\epsilon +2\epsilon +2\epsilon =6\epsilon$. As $\epsilon >0$ was arbitrary and $\mathrm{\mid }\mathrm{\Sigma }R-S\mathrm{\mid }\ge 0$, this forces $\mathrm{\Sigma }R=S$, 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 $a_{J^{\mathsf{T}}(n)}^{\mathsf{T}}=a_{J(n)}$ for every $n$, so the two linear series are the same series and have the same sum $S$.