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I have some questions concerning convergence of sums where the summands are complex number, although the real motivation of my question comes from Von Neumann algebras where sometimes the summands are more general.

Background: There are two competing notions of infinite sums. One is over indexing from the natural numbers, where convergence is as usual and one can also speak of absolute convergence. The other is indexing from an arbitrary set $S$, and convergence is defined by using the directed set of finite subsets of the arbitrary set $S$, as ordered by inclusion, and the net elements are the sums over the finite subsets of $S$. I will refer to the former as the sequential sum, and the latter as the net sum. Some background facts that may be useful are as follows:

  1. A sequential sum is absolutely convergent if and only if the value and convergence of the sum is independent of permutations.
  2. A net sum is convergent if and only if only countably many summands are nonzero, and those summands are l^1 or equivalently absolutely convergent as a sequential sum.
  3. A net sum with indexing by the natural numbers is convergent if and only if it is absolutely convergent when viewed as a sequence sum.

My question is suppose that $I$ and $J$ are index sets, hence we can also consider their cartesian product as an index set, for which it makes sense to speak of the net sum. Suppose I have an associated set of numbers $a_{ij}$ for each $i$ in $I$ and $j$ in $J$. When can I conclude that I can compute the joint sum over $I$ and $J$ of this set of numbers by computing a sum in $I$ first and then in $J$ or vice versa? By Fubini's theorem, if $I$ and $J$ are sigma-finite wrt counting measure, which can be arranged for by reducing to such a case via observation 2., then we can do this when the joint sum is convergent, since then it would also be absolutely convergent by observation 2. But I need better than this for my applications. I need to know the following:

Question 1: if $\sum_{i}\sum_{j}a_{ij}=\sum_{j}\sum_{i}a_{ij}$ (both sums both sides being convergent as nets) then $\sum_{ij}a_{ij}$ converges to the same value where the index set is $I\times J$

Please let me know if this is true and how to prove your answer. It is false for the sequential notion of sum, but the discussion above shows that net sums come equipped already with absolute convergence. The problem is that if one simply uses observation 2, it tells me that in the equality above, each side has the inner sum being absolutely convergent, and the outer sum being absolutely convergent, but not that I can slip the absolute value bars induced by the outer sum through the inner sum.

Question 2: Actually, one can really consider net sums in a topological vector space, although sequence sums stop being such a desirable topic. Often times, for instance in the case of the Strong Operator topology, one simply writes sums over two indices i and j as if it's unimportant whether I do i first, then j, j first, then i, or i and j together. However, whenever I verify their arguments, I can only do it at best usually for i first then j, or j first then i, and my arguments break down when it comes down to doing a joint sum. Most often, this occurs with things like saying if $x$ is a bounded operator and $p_i$ is a decomposition of unity into projections in $B(H)$ then $x=\sum_{ij}p_jxp_i$ where as you can understand I have trouble showing convergence because I don't think it's true, although it is clearly true if you replace the joint sum with separate sums in either order. Are there hidden assumptions or arguments here that warrant this level of carefreeness? I even had a more advanced graduate student tell me that what is usually meant is the joint sum, but I am not certain of how correct that is.

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Okay, for some reason all my comments were moved to be under a previously existing answer, which was then deleted. In any case, the important things that were stated besides corrections to the answer was just the counterexample to claim 1. Just let $a_{ij}$ be 2 on the main diagonal, and -1 on the super and subdiagonal. – Jeff Jun 29 '12 at 16:14

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