The array with $a_{ii} = 1$, $a_{i+1,i} = -1$ and every other entry $0$ has iterated sums $1$ and $0$

Define $a : \mathbb{N}\times\mathbb{N} \to \mathbb{R}$ by

$a_{ij} := \begin{cases} 1 & \text{if } j = i, \\ -1 & \text{if } i = j+1, \\ 0 & \text{otherwise,} \end{cases}$

so that the array has $1$ along the diagonal, $-1$ immediately below it, and $0$ everywhere else. Then every row series and every column series converges, both series of those sums converge, and

$\sum_{i=0}^{\infty}\Bigl(\sum_{j=0}^{\infty} a_{ij}\Bigr) \;=\; 1, \qquad \sum_{j=0}^{\infty}\Bigl(\sum_{i=0}^{\infty} a_{ij}\Bigr) \;=\; 0 .$

The two iterated sums exist and differ, which is FALSE: whenever both iterated sums of a double array exist, they are equal. What this example adds is the reason 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 does not apply: the row totals of absolute values are $A_0 = 1$ and $A_i = 2$ for every $i \ge 1$, so $\sum_i A_i$ diverges and the hypothesis of that theorem fails at its only substantive point.

Written out, the array is

$\begin{array}{cccc} 1 & 0 & 0 & \cdots \\ -1 & 1 & 0 & \cdots \\ 0 & -1 & 1 & \cdots \\ 0 & 0 & -1 & \cdots \end{array}$

Every row after the first contains one $+1$ and one $-1$ and so sums to $0$; every column contains one $+1$ and one $-1$ and so sums to $0$. The asymmetry is that the very first row has no $-1$ to its left, and that single missing entry is the whole difference between $1$ and $0$.

• The array is a discrete telescoping, and one entry falls off the edge. Each $-1$ sits one row below the $+1$ of its own column: the $+1$ at $(j,j)$ and the $-1$ at $(j+1,j)$ are both in column $j$, so every column cancels within itself and every column total is $0$. Along a row the same two entries of that row, $a_{ii} = +1$ and $a_{i,i-1} = -1$, also cancel — but only for $i \ge 1$. Row $0$ has no $-1$, there being no column $-1$ for it to lie in, so its total is the uncancelled $+1$. That single missing entry, and nothing else, is the whole difference between the two iterated sums. Which cancellation one sees depends only on the order of summation.

• Nothing here is large. Every entry is $0$, $1$ or $-1$, every row and column has at most two nonzero entries, and the discrepancy between the two iterated sums is exactly $1$. The failure of Fubini's theorem without an absolute hypothesis is not a phenomenon of large or wild arrays.

• The connection with rearrangement is exact. By 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, under the absolute hypothesis the common value is also the sum along any enumeration of $\mathbb{N}\times\mathbb{N}$; here no such common value exists, just as a conditionally convergent series has no order-independent sum (The Riemann series theorem: a conditionally convergent real series has, for every $c \in \mathbb{R}$, a rearrangement with sum $c$, and rearrangements diverging to $+\infty$, to $-\infty$, and oscillating with any prescribed $\liminf \le \limsup$ in $\overline{\mathbb{R}}$).