Double Series and Fubini

Numbers on a grid get summed by rows first, then by columns, in this scene. With absolute convergence, the two answers agree.

Add along the rows or down the columns: the result is one and the same. The order cannot change it because the absolute values sum to a finite total. With only conditional convergence, swapping the order changes the answer itself. The headline value is the second formula below.

For example, stack 📦 boxes in a grid and count the items inside each box. Count row by row or column by column — the total count comes out identical.

Examples & Exercises

Prove Fubini for infinite sums (nonnegative terms) completely: if $a_{m,n} \ge 0$ then $\sum_m \sum_n a_{m,n} = \sum_n \sum_m a_{m,n}$ .

Show both sides equal the supremum $S$ of sums over finite rectangles.

S = \sup_{M,N} \sum_{m \le M} \sum_{n \le N} a_{m,n} \in [0, \infty]

\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n} = S = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_{m,n}

N = 5

: sum by rows

\approx 2.1422

, sum by columns

\approx 2.1422

— both head to

(\pi^2/6)^2 = \pi^4/36 \approx 2.7058

Show proof

Put $T(M,N) = \sum_{m \le M}\sum_{n \le N} a_{m,n}$ (a finite sum, $\ge 0$ ) and $S = \sup_{M,N} T(M,N) \in [0,\infty]$ .
$T$ is monotone in each of $M, N$ .
For each fixed $m$ , the inner series $\sum_{n=1}^{\infty} a_{m,n} \in [0,\infty]$ converges (finitely or to $+\infty$ ) since the terms are nonnegative.
Put $R = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n}$ .
For each $M, N$ : $T(M,N) = \sum_{m \le M}\sum_{n \le N} a_{m,n} \le \sum_{m \le M}\sum_{n=1}^{\infty} a_{m,n} \le R$ (inner finite sum $\le$ inner infinite sum, outer finite $\le$ outer infinite — all terms nonnegative).
Taking the supremum over $M, N$ gives $S \le R$ .
The other direction: for each $M$ , $\sum_{m \le M}\sum_{n=1}^{\infty} a_{m,n} = \sup_N T(M,N) \le S$ (the inner infinite sum is the supremum of the inner finite sums).
This equality needs justification: for fixed $M$ , each inner sum $\sum_{n \le N} a_{m,n}$ is increasing in $N$ (since $a_{m,n} \ge 0$ ) and converges to $\sum_{n=1}^{\infty} a_{m,n}$ .
For a finite sum ( $M$ terms), supremum and summation interchange — a finite sum of increasing sequences is increasing, and its supremum equals the sum of the suprema (checked termwise).
Hence $\sum_{m \le M} \sup_N \sum_{n \le N} a_{m,n} = \sup_N \sum_{m \le M} \sum_{n \le N} a_{m,n} = \sup_N T(M,N)$ .
Taking the supremum over $M$ gives $R = \sup_M \sum_{m \le M}\sum_{n=1}^{\infty} a_{m,n} \le S$ .
Hence $R = S$ .
Swapping the roles of rows and columns, the identical argument gives $C = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_{m,n} = S$ (the same supremum of the same $T(M,N)$ ).
Hence $R = C = S$ — the two iterated sums agree.
If $S < \infty$ the common value is finite; if $S = \infty$ both sides diverge to $+\infty$ .
QED.

Extend to the absolutely convergent case (short complete proof): if $\sum_{m,n} \lvert a_{m,n}\rvert < \infty$ , use the decomposition $a = a^+ - a^-$ to apply the first problem to each part and conclude $\sum_m\sum_n a_{m,n} = \sum_n\sum_m a_{m,n}$ .

a^+ = \max(a, 0), \quad a^- = \max(-a, 0), \quad \lvert a\rvert = a^+ + a^-, \quad a = a^+ - a^-

\sum_m\sum_n a = \sum_m\sum_n a^+ - \sum_m\sum_n a^- = \sum_n\sum_m a^+ - \sum_n\sum_m a^- = \sum_n\sum_m a

N = 5

: signs are present but

\sum \lvert a\rvert

converges, so sum by rows

\approx 0.7033

and sum by columns

\approx 0.7033

agree, both heading to

\pi^4/144 \approx 0.6765

Show proof

Put $a^+_{m,n} = \max(a_{m,n}, 0)$ and $a^-_{m,n} = \max(-a_{m,n}, 0)$ .
Then $0 \le a^{\pm}_{m,n} \le \lvert a_{m,n}\rvert$ , with $\lvert a\rvert = a^+ + a^-$ and $a = a^+ - a^-$ .
By comparison, both $\sum a^+_{m,n}$ and $\sum a^-_{m,n}$ converge, bounded by $\sum \lvert a_{m,n}\rvert < \infty$ .
Since $a^+$ and $a^-$ are nonnegative, apply the first problem (Fubini, nonnegative terms) to each: $\sum_m\sum_n a^+ = \sum_n\sum_m a^+ < \infty$ , and likewise for $a^-$ .
All iterated sums are finite, so subtraction is legitimate, and $\sum_m\sum_n a = \sum_m\sum_n a^+ - \sum_m\sum_n a^- = \sum_n\sum_m a^+ - \sum_n\sum_m a^- = \sum_n\sum_m a$ .
QED.

Show the hypothesis is sharp: with $a_{n,n} = 1$ , $a_{n,n+1} = -1$ ( $n \ge 1$ ) and $0$ elsewhere, verify completely that every row sum is $0$ (so $\sum_m\sum_n a_{m,n} = 0$ ) but column $1$ sums to $1$ and columns $n \ge 2$ sum to $0$ (so $\sum_n\sum_m a_{m,n} = 1$ ), and note $\sum \lvert a\rvert = \infty$ .

a_{n,n} = 1, \quad a_{n,n+1} = -1 \ (n \ge 1), \quad \text{else } 0

\sum_m \sum_n a_{m,n} = \sum_m 0 = 0 \quad\ne\quad 1 = 1 + \sum_{n \ge 2} 0 = \sum_n \sum_m a_{m,n}

The alternating series in the picture above is absolutely convergent, so reordering is safe — but as the counterexample below shows, with

\sum\lvert a\rvert = \infty

the order can change the answer.

Show proof

Fix a row ( $m$ fixed).
The only nonzero entries are $a_{m,m} = 1$ at $n = m$ and $a_{m,m+1} = -1$ at $n = m+1$ .
Hence the row sum is $\sum_{n=1}^{\infty} a_{m,n} = 1 + (-1) = 0$ (a finite sum, so no convergence issue).
This holds for every $m$ , so $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n} = \sum_{m=1}^{\infty} 0 = 0$ .
Fix a column ( $n$ fixed).
The nonzero entries: $a_{n,n} = 1$ (row $m = n$ ) and, for $n \ge 2$ , $a_{n-1,n} = -1$ (row $m = n-1$ , from $a_{m,m+1} = -1$ with $m = n-1$ ).
Column $1$ : only $a_{1,1} = 1$ , so the sum is $1$ .
Column $n \ge 2$ : $1 + (-1) = 0$ .
Hence $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_{m,n} = 1 + \sum_{n=2}^{\infty} 0 = 1$ .
The two iterated sums differ, $0 \ne 1$ — the order changes the answer.
The broken hypothesis is absolute convergence: $\sum_{m,n} \lvert a_{m,n}\rvert = \sum_{n=1}^{\infty} (1 + 1) = \infty$ , so the second problem's hypothesis fails.
Without the nonnegative (or absolutely convergent) hypothesis Fubini fails, and this example shows the hypothesis is sharp.
QED.