TITLE: Gauge Integral: well defined?
QUESTION [6 upvotes]: Given a compact space $\Omega$ and a Banach space $E$
Consider functions $f:\Omega\to E$.
Regard neighborhood gauges:
$$\delta:\Omega\to\mathcal{T}(\Omega):\quad\delta(\omega)\in\mathcal{N}(\omega)$$
and finite measurable tagged partitions:
$$\mathcal{P}^*\subseteq\mathcal{B}(\Omega):\quad\#\mathcal{P}^*<\infty$$
(In fact, the tags are just surpressed.)
Order gauges by inclusion:
$$\delta\leq\delta':\iff\delta(\omega)\subseteq\delta'(\omega)\quad(\omega\in\Omega)$$
and collect gauge-fine partitions:
$$\mathcal{P}^*\dashv\delta:\iff A_n\subseteq\delta(a_n)\quad(A_n\in\mathcal{P}^*)$$
Denote the partial sums by:
$$\mathcal{S}(\mathcal{P}^*)=\sum_nF(a_n)\lambda(A_n)$$
and define the gauge integral to be the limit:
$$\int_\Omega F\mathrm{d}\lambda:=\lim_\delta\{\mathcal{S}(\mathcal{P}^*)\}_{\mathcal{P}^*\dashv\delta}$$
Why is the value assigned to a gauge integral well defined (unique)?

REPLY [3 votes]: Gauges are directed:
$$\delta\wedge\delta'(\omega):=\delta(\omega)\cap\delta'(\omega)$$
and every gauge admits a partition:
$$\Omega=\bigcup_{n=1}^N\delta(a_n)=:\bigcup_{n=1}^NA_n:\quad\mathcal{P}_\delta:=\{A_1',\ldots,A_N'\}\quad(A_n':=A_n\setminus\cup_{m=1}^{n-1}A_m)$$
(Note that compactness was necessary!)
That is gauges induce a net of partial sums:
$$\mathcal{S}(\mathcal{P}_\delta^*):\quad{\mathcal{P}}_{\delta}^{*}\leq{\mathcal{P}'}_{\delta'}^{*'}:\iff\delta\leq\delta'$$
(Caution: There may be no meet!)
Concluding that the integral is well-defined in Banach space $E$.