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In our analysis course, the following question came up and could, up to now, not be solved:

Let $a: \mathbb{N} \to \mathbb{C}$ be a sequence of complex numbers. What are necessary and sufficient conditions for the existence of a function $\mu: 2^\mathbb{N} \to \mathbb{C}$ satisfying the properties

  • $\mu(\{i\}) = a_i$
  • $\mu$ is finitely additive, i.e. $A\cap B = \emptyset\implies \mu(A\cup B) = \mu(A)+\mu(B)$?

Partial Results

If we are given such a function defined on a subset of $2^\mathbb{N}$, we can of course extend it to all sets of the form $A\cup B, A\cap B = \emptyset$ and $A\setminus B, B\subset A$. This invites the use of Zorn's Lemma; but it seems impossible to prove that a maximal set closed under these operations must be $2^\mathbb{N}$. However, this approach strongly suggests that $\mu$ exists for all $(a_i)$, as the problem only depends on $2^\mathbb{N}$.

On the other hand, if $a_i$ converges absolutely, one can set $\mu(I) = \sum_{i\in I} a_i$ which fulfills the required properties, but this approach does not generalise at all.

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Very nice question. It doesn't seem that complex numbers have much to do with it, since the problem splits neatly into the real and imaginary parts. Similarly, it seems you might as well assume all $a_i$ are positive real numbers, since again the solution splits into separate solutions for the positive and negative terms. – JDH Dec 16 '10 at 16:02
But then, you can just use Caratheodory's extension theorem, can't you? – Stefan Dec 16 '10 at 16:10
With Caratheodory, you would get infinite values, if the series does not converge, but the OP wants all values finite. – JDH Dec 16 '10 at 16:23
up vote 8 down vote accepted

Given any such sequence $a_i$, you can use it to define a finitely additive measure on the collection of finite subsets of $\mathbb N$. Now consider the vector space $V:= \mathbb C^{\oplus \mathbb N}$, i.e. the direct sum of countably many copies of $\mathbb C$, with the copies of $\mathbb C$ being indexed by elements $i \in \mathbb N$. Alternatively, this is the space of sequences $(z_i)_{i \in \mathbb N}$ with $z_i =0$ for all but finitely many $i$. Or, if you want to think a little more analytically, you can think of this as the space of $\mathbb C$-valued functions on $\mathbb N$ with finite support, i.e. which vanish outside a finite set. (The function attached to a sequence is just $i \mapsto z_i$, of course.)

Our choice of finitely addivite measure defines a functional on $V$, given by $(z_i) \mapsto \sum_{i \in \mathbb N} a_i z_i.$ (More analytically, this is integration of the finitely supported function corresponding to $(z_i)$ against our finitely addivite measure.)

Now let $W$ be the vector space of all $\mathbb C$-valued functions on $V$. Certainly $V \subset W$, and we may always extend linear functionals from a subspace to the whole space; thus we may extend our given functional to a functional $I: W \to \mathbb C$. (The label $I$ is chosen to suggest integration.)

Now if $S$ is any subset of $\mathbb N$, let $\chi_S$ be the characteristic function of $S$. Define $\mu(S) = I(\chi_S)$. Then $\mu$ is a finitely additive measure on $\mathbb N$ satisfying the required properties.

(This is essentially the Zorn's lemma argument suggested in the original posting, but reformulated in terms of extending functionals on vector spaces, which makes it more transparent.)

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This has a similar feeling to the construction of a Banach limit ( ), except that the desired conditions on the functional being generated here are different than the conditions on the functional being generated in that context. – Carl Mummert Dec 16 '10 at 16:30
@Carl: Dear Carl, I agree, and in fact that is where I got the idea for this construction. Regards, – Matt E Dec 16 '10 at 16:45
Nice answer! I'm still thinking why this application of Zorn's Lemma does not run into the same problems as the one I suggested, but it seems that complex linear independence is much better behaved than closedness under disjoint unions and removal of subsets. – Bertram Dec 17 '10 at 7:44

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