Generalized Riemann Integral: Improper Version


For a bounded nonexample of integrability see: Riemann Integral: Bounded Nonexample

For a convergence theorem on integral see: Riemann Integral: Uniform Convergence

For a comparison of integrals see: Uniform Integral vs. Riemann Integral


Given a measure space $\Omega$ and a Banach space $E$.

Consider functions $F:\Omega\to E$.

Denote the measurable subsets of finite mass by:
and order them by inclusion:
$$A\leq A’:\iff A\subseteq A’$$

Remember the generalized Riemann integral on finite measure spaces:
$$A\in\mathcal{A}_\infty:\quad\int_AF\mathrm{d}\mu:=\lim_\mathcal{P}\left\{\sum_{a\in A\in\mathcal{P}}F(a)\mu(A)\right\}_\mathcal{P}$$
(For more details see references above.)

Define the improper Riemann integral as:
$$\int_\Omega F\mathrm{d}\mu:=\lim_A\left\{\int_AF\mathrm{d}\mu\right\}_{A\in\mathcal{A}_\infty}$$
(Crucially, this reflects independence of approximation by finite spaces.)


For finite measure spaces the improper agrees with the proper as $\Omega\in\mathcal{A}_\infty$.

This way, poles still can’t be handled:
$$\int_0^1\frac{1}{\sqrt{x}}\mathrm{d}x\notin E$$
(Note that the concept of compact intervals isn’t available in general.)

For Borel spaces a suitable criterion could be continuity plus absolute integrability:
$$F\in\mathcal{C}(\Omega,E):\quad\int_\Omega\|F\|\mathrm{d}\mu<\infty\implies\int_\Omega F\mathrm{d}\mu\in E$$

How to prove this in the abstract setting?

(I slightly doubt it…)

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