Interchanging the Order of Nonnegative Double Sums

Why the Order of Summation Is a Question

A double array can be summed in two natural ways: add each row and then add the row totals, or add each column and then add the column totals. For nonnegative terms, these procedures always agree, even when their common value is infinite. The reason is not a rearrangement rule for arbitrary series: it is that both procedures amount to taking the supremum of the same finite sums.

Throughout, indices \(m,n\) range over \(\mathbb{N}=\{1,2,\ldots\}\). For a real-valued array \(a_{mn}\ge0\), define a single series as the limit of its increasing partial sums, allowing the value \(+\infty\). The array can also be viewed as a measurable function on \((\mathbb{N}^2,\mathcal{P}(\mathbb{N}^2),\#)\), where \(\#\) is counting measure. This measure space is complete and sigma-finite, but not finite: \(\mathbb{N}^2\) is a countable union of finite-measure singleton sets and has infinite measure. Its integral of a nonnegative function is the corresponding sum.

For positive integers \(M,N\), write \(R_{M,N}=\sum_{m=1}^{M}\sum_{n=1}^{N}a_{mn}\). These rectangular sums increase when either side of the rectangle increases. The key fact is that each iterated sum equals the supremum of all such rectangles.

Finite Rectangles Determine Either Iterated Sum

Proof. For every \(n\), \(x_{i,n}\le\sup_jx_{i,j}\) for each \(i\), so \(\sup_n\sum_{i=1}^k x_{i,n}\le\sum_{i=1}^k\sup_jx_{i,j}\). For the reverse inequality, if any \(\sup_nx_{i,n}=+\infty\), then \(\sup_n\sum_i x_{i,n}=+\infty\), since that sum is at least \(x_{i,n}\) for every \(n\). Otherwise all \(k\) suprema are finite. Given \(\varepsilon>0\), for each \(i\) choose \(n_i\) with \(x_{i,n_i}>\sup_n x_{i,n}-\varepsilon/k\). Let \(n=\max_i n_i\). Monotonicity gives \(x_{i,n}\ge x_{i,n_i}\) for every \(i\), and hence \(\sum_i x_{i,n}>\sum_i\sup_jx_{i,j}-\varepsilon\). Since this holds for every \(\varepsilon>0\), the reverse inequality follows. \(\square\)

Proof. Set \(r_m=\sum_{n=1}^{\infty}a_{mn}=\sup_N\sum_{n=1}^{N}a_{mn}\). By the definition of the outer nonnegative series, the row-first sum is \(\sup_M\sum_{m=1}^{M}r_m\). Fix \(M\). Apply the lemma to the \(M\) nondecreasing sequences \(x_{m,N}=\sum_{n=1}^{N}a_{mn}\). It gives \(\sum_{m=1}^{M}r_m=\sup_N\sum_{m=1}^{M}\sum_{n=1}^{N}a_{mn}=\sup_NR_{M,N}\). Taking the supremum over \(M\), the row-first sum is \(\sup_{M,N}R_{M,N}\).

For the column-first sum, set \(c_n=\sum_{m=1}^{\infty}a_{mn}\). For fixed \(N\), apply the lemma to the \(N\) nondecreasing sequences \(\sum_{m=1}^{M}a_{mn}\). This gives \(\sum_{n=1}^{N}c_n=\sup_M R_{M,N}\). Taking the supremum over \(N\), the column-first sum is again \(\sup_{M,N}R_{M,N}\). Thus both sums are equal, including when this supremum is infinite. \(\square\)

This proof uses only nonnegativity and the definitions of the sums. Equivalently, on the stated counting-measure space, the equality follows by applying the Monotone Convergence Theorem to finite partial sums. That theorem applies because the relevant partial sums are measurable, nonnegative, and increase pointwise. The finite-rectangle proof makes the common value explicit.

Worked Calculations

Where Nonnegativity Matters

Nonnegativity ensures that enlarging a finite rectangle cannot reduce its sum and that each iterated series is an increasing limit. Without that condition, the rectangle argument does not apply: cancellation can make the two orders disagree.

For a concrete example on indices \(m,n\ge1\), let \(b_{mn}=1\) when \(n=m\), let \(b_{mn}=-1\) when \(n=m+1\), and let \(b_{mn}=0\) otherwise. Each row has exactly two nonzero entries, \(1\) and \(-1\), so its sum is \(0\); summing row totals gives \(0\). In column \(n=1\), the only nonzero entry is \(b_{11}=1\), so that column sums to \(1\). For every \(n\ge2\), the nonzero entries in column \(n\) are \(b_{nn}=1\) and \(b_{n-1,n}=-1\), so its sum is \(0\). Summing the column totals therefore gives \(1\), not \(0\).

This is not a counterexample to the theorem: \(b_{mn}\) is signed, not nonnegative. Indeed, the sum of absolute values in every row is \(2\), so the sum of all absolute values is infinite. The theorem cannot be used to interchange these signed sums.

A common mistake is to apply the nonnegative result to an array with negative entries and then conclude that the iterated sums must agree. The calculation above shows exactly what fails: the row-first value is \(0\), while the column-first value is \(1\). Before interchanging an order, check that every term is nonnegative, or establish an appropriate hypothesis for signed-series interchange; the nonnegative theorem alone supplies no such conclusion.