Q: Generalized Riemann Integral: Uniform Convergence Disclaimer
This thread is meant to record. See: Answer own Question
And it is written as question. Have fun! :)
Reference
For a bounded nonexample of integrability see: Riemann Integral: Bounded Nonexample
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 $E$.
(In fact, a Hausdorff TVS should be sufficient.)
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.)
Denote Riemann integrable functions by $\mathcal{L}(\mu)$.
Problem
Does a.e. uniform convergence imply integrability and allow interchange of limits:
$$F_n\in\mathcal{L}(\mu):\quad F_n\to F\implies\int F_n\mathrm{d}\mu\to\int F\mathrm{d}\mu\quad(F\in\mathcal{L}(\mu))$$
(Although this seems to trivial it needs a proof anyway.)
Thus the Riemann integral includes the uniform integral: $\mathcal{L}_\mathfrak{U}\subseteq\mathcal{L}_\mathfrak{R}$

A: First of all, check integrability:
$$\mathcal{P},\mathcal{P}'\geq\mathcal{P}_{N(\varepsilon)}:\quad\|\sum_{\mathcal{P}}F(a)\mu(A)-\sum_{\mathcal{P}'}F(a')\mu(A')\|\\\leq\|F-F_N\|_\infty\mu(\Omega)
+\|\sum_{\mathcal{P}}F_N(a)\mu(A)-\sum_{\mathcal{P}'}F_N(a')\mu(A')\|+\|F-F_N\|_\infty\mu(\Omega)<3\left(\frac{\varepsilon}{3}\right)$$
Next, check convergence:
$$n\geq N(\varepsilon)\in\mathbb{N}:\quad\|\int F\mathrm{d}\mu-\int F_n\mathrm{d}\mu\|\\\leq\|\int F\mathrm{d}\mu-\sum_{\mathcal{P}_n}F(a)\mu(A)\|+\|F-F_n\|_\infty\mu(\Omega)+\|\int F_n\mathrm{d}\mu-\sum_{\mathcal{P}_n}F_n(a)\mu(A)\|<3\left(\frac{\varepsilon}{3}\right)$$
