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(Assume Countable Choice.)

Generalizing and,
the Riemann-Lebesgue integral can be defined as follows:

For $V$ a real $(\text{T}_0)$ topological vector space, $\Omega$ the underlying set of a measure space with measure $\mu$,

$\: f : \Omega \to V \:$ a function, and $v$ a vector, $\;\; \displaystyle\int f \: = \: v \;\;$ if and only if

for all open subsets $U$ of $V$, if $\:v\in U\:$ then there exists a countable partition $\Gamma\hspace{0.01 in}'$ of $\:\Omega\:$ into non-empty measurable pieces such that [[for all members $x$ of $\Omega$, if $x\hspace{0.02 in}$'s piece has infinite measure then $\:f(x)$
is the zero vector] and [for all countable refinements $\Gamma$ of $\Gamma\hspace{0.01 in}'$, for all choice functions $\: c : \Gamma \to \Omega \:$, $\:$ $\: \displaystyle\sum_{S\in \Gamma}\:(\mu(S) \cdot f(c(S))) \:$ converges unconditionally to an element of $U$]].

Since the definitions provided in both those sources do use refinements and not just $\Gamma\hspace{0.01 in}'$, presumably something would go wrong otherwise. $\:$ I notice that the only way something can go wrong from that is if uniqueness breaks. $\:$ I'm also wondering what difference not requiring the sums to converge might make.

${}$1. $\:$ What (if anything) is an example where uniqueness would
fail if the choice functions and sums used $\Gamma\hspace{0.01 in}'$ instead of $\Gamma$?

${}$2. $\:$ Would one actually get something different if one replaced the last line of the definition with:
there exists a finite subset $\Gamma_0$ of $\Gamma$ such that for all finite subsets $\Gamma_1$ of $\Gamma$,
if $\: \Gamma_0 \subseteq \Gamma_1 \;$ then $\;\;\;\; \displaystyle\sum_{S\in \Gamma_1}\:(\mu(S) \cdot f(c(S))) \;\; \in \;\; U \;\;\;\;$?

${}$3. $\:$ Does assuming that $V$ is locally convex change the answers to 1 or 2?

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I think this is more appropriate here than at math.stackexchange, but I might be wrong. $\hspace{0.5 in}$ – Ricky Demer Oct 3 '12 at 7:08

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