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: $$\mathcal{A}_\infty:=\{A:\mu(A)<\infty\}$$ 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...)


Yes, it holds!

As it is continuous it is Bochner measurable by Pettis's criterion.

As it is absolutely integrable it is also Bochner integrable.

But it is bounded so on subspaces of finite measure Riemann integrable.

Thus by dominated convergence also improperly Riemann integrable.


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