Generalized Riemann Integral: Bounded Nonexample?

Reference

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

For an improper version of integral see: Riemann Integral: Improper Version

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

Definition

Given a finite measure space $\mu(\Omega)<\infty$ and a Banach space $E$.

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

Define the generalized Riemann integral by: $$\int F\mathrm{d}\mu:=\lim_\mathcal{P}\{\sum_{a\in A\in\mathcal{P}}F(a)\mu(A)\}_\mathcal{P}$$ over finite measurable partitions: $$\mathcal{P}\subseteq\Sigma:\quad\Omega=\bigsqcup_{A\in\mathcal{P}}A\quad(\#\mathcal{P}<\infty)$$ being ordered by refinement: $$\mathcal{P}\leq\mathcal{P}':\iff\forall A'\in\mathcal{P}'\exists A\in\mathcal{P}:\quad A\supseteq A'$$ (In fact, the tags are just surpressed.)

Discussion

What do nonintegrable functions look like? $$F:\Omega\to E:\quad F\notin\mathcal{L}_\mathfrak{R}(\mu)$$ Obviously, nonmeasurable nonexamples are countless: $$\mu^*(V)\neq\mu_*(V):\quad F=\chi_V$$ Also, a.e. boundedness is a necessary condition on integrability: $$F\notin\mathcal{L}^\infty(\mu)\implies F\notin\mathcal{L}_\mathfrak{R}(\mu)$$ So are there bounded measurable nonexamples?
(Note that continuity is not an accessible concept here.)

Besides, pathological examples like the Dirichlet function become integrable.

Deep Consequence

Riemann integrability is not guaranted by absolute integrabilit alone: $$\int\|F\|\mathrm{d}\mu<\infty\nRightarrow \int F\mathrm{d}\mu\in E$$ But this shouldn't be to big surprise as this wasn't either the case for Bochner integrability.

Solution 1:

Ok, I think I got it now...

Take a slight variation of the famous example: $$F:[0,1]\to\ell[0,1]:t\mapsto\chi_t$$

At least, it is bounded: $\|F(t)\|\equiv1$.

Especially, it is absolutely integrable: $\int\|F(t)\|\mathrm{d}t=1$

However, it is not measurable as: $$\|\chi_s-\chi_t\|^2=2\quad(s\neq t)$$ so taking a Vitali set yields: $$U:=\bigcup_{v\in V}B_{\frac{1}{\sqrt{2}}}(\chi_v):\quad U\in\mathcal{T}[0,1],F^{-1}U=V\notin\mathcal{B}[0,1]\quad\left(V\subseteq[0,1]\right)$$

Moreover, it has nonseparable range: $$\|\chi_s-\chi_t\|\equiv2\quad(s\neq t)$$ So it can't be pointwise limit in any case: $S_n\nrightarrow F$

Finally, it is not Riemann integrable as: $$\|\sum_{A\in\mathcal{P}}F(a)\mu(A)-\sum_{A\in\mathcal{P}}F(a')\mu(A)\|=\sum_{A\in\mathcal{P}}\|\chi_a-\chi_{a'}\|\mu(A)=2\sum_{A\in\mathcal{P}}\mu(A)\equiv2\mu(\Omega)\quad(a\neq a')$$