Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(Assume Countable Choice.)

Generalizing arxiv.org/pdf/math/9903103 and www.um.es/beca/papers/BirkhoffBourgain.pdf,
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?

share|improve this question
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.