Bochner Integral vs. Riemann Integral Disclaimer
This thread is meant to record. See: Answer own Question
Anyway, it is written as problem. Have fun! :)
Reference
This thread is directly related to:


*

*Bochner Integral: Axioms

*Bochner Integral: Integrability

*Riemann Integral: Bounded Nonexample

*Riemann Integral: Uniform Convergence

*Riemann Integral: Improper Version

*Riemann Integral vs. Uniform Integral
and it is indirectly related to:


*

*Stone's Theorem Integral

*Nonmeasurable Functions

*Pointwise Limit vs. Uniform Limit
Prologue
This is the final comparison.
All notions are up to null sets.
Definitions
Given a finite measure space $\mu(\Omega)<\infty$ and a Banach space $E$.
Consider plain functions $F:\Omega\to E$.
Denote the Bochner measurable functions by $\mathcal{B}$.
Define the Bochner integral by:
$$F\in\mathcal{B}:\quad\int F\mathrm{d}\mu:=\lim_n\int S_n\mathrm{d}\mu$$
for simple functions satisfying:
$$\int\|F-S_n\|\mathrm{d}\mu\to0$$
Denote the Bochner integrable functions by $\mathcal{L}_\mathfrak{B}$.
Define the generalized Riemann integral by:
$$\int_\mathfrak{R}F\mathrm{d}\mu:=\lim_\mathcal{P}\{\sum_{a\in A\in\mathcal{P}}F(a)\mu(A)\}_\mathcal{P}$$
over finite measurable partitions:
$$\mathcal{P}\subseteq\Sigma:\quad\Omega=\bigsqcup_{A\in\mathcal{P}}A\quad(\#\mathcal{P}<\infty)$$
being ordered by refinement:
$$\mathcal{P}\leq\mathcal{P}':\iff\forall A'\in\mathcal{P}'\exists A\in\mathcal{P}:\quad A\supseteq A'$$
Denote Riemann integrable functions by $\mathcal{L}_\mathfrak{R}$.
Problem
Clearly, boundedness is a necessary condition for Riemann integrability:
$$f:(0,1]\to\mathbb{R}:x\mapsto\frac{1}{\sqrt{x}}:\quad f\in\mathcal{L}_\mathfrak{B},\,f\notin\mathcal{L}_\mathfrak{R}$$
However, there are also bounded Riemann integrable functions that are not Bochner integrable:
$$F:[0,1]\to\ell^2[0,1]:t\mapsto\chi_t:\quad F\notin\mathcal{L}_\mathfrak{B},\,F\in\mathcal{L}_\mathfrak{R}$$
Moreover, both strictly include the uniform integral:
$$\mathcal{L}_\mathfrak{U}\subsetneq\mathcal{L}_\mathfrak{B}\land\mathcal{L}_\mathfrak{U}\subsetneq\mathcal{L}_\mathfrak{R}$$

So the remaining question is wether at least for the bounded case:
  $$\|F\|_\infty<\infty:\quad F\in\mathcal{L}_\mathfrak{B}\implies F\in\mathcal{L}_\mathfrak{R}$$
  and especially:
  $$\int_\mathfrak{R}F\mathrm{d}\mu=\int_\mathfrak{B}F\mathrm{d}\mu$$
  (Have a guess! ;))

 A: Proof
Consider a bounded function $\|F\|_\infty<\infty$.
Now, by measurability there exists a simple approximation:
$$S_n\in\mathcal{S}:\quad\|F-S_n\|\to0$$
After cutoff and reset it can be chosen to be bounded and decreasing:
$$\|S_n\|\leq2\|F\|:\quad\|F-S_n\|\downarrow0$$
Next, denote sufficiently close points by:
$$\Omega_n:=\{\|F-S_n\|<\varepsilon\}$$
These are increasing so by continuity from above:
$$\mu(\Omega)<\infty:\quad\mu(\Omega_n^\complement)\stackrel{n\to\infty}{\to}0$$
Finally, choose a partition where simple functions are constant:
$$S_n=\sum_{k=1}^KS_k\chi(A_k):\quad\mathcal{P}_n:=\{A_1,\ldots,A_K\}$$
Putting all together one obtains the estimate:
$$0\leq\|\sum_{\mathcal{P}}F(a)\mu(A)-\sum_{\mathcal{P}'}F(a')\mu(A')\|\leq\left(\|F-S_N\|_{\Omega_N}\mu(\Omega_N)+\|F-S_N\|_{\Omega_N^\complement}\mu(\Omega_N^\complement)\right)\\+\|S_N-S_N\|_\infty\mu(\Omega)+\left(\|F-S_N\|_{\Omega_N}\mu(\Omega_N)+\|F-S_N\|_{\Omega_N^\complement}\mu(\Omega_N^\complement)\right)\\\leq2\left(\|F-S_N\|_{\Omega_N}\mu(\Omega)+3\|F\|_\infty\mu(\Omega_N^\complement)\right)+0<\varepsilon\quad\left(\mathcal{P},\mathcal{P}'\geq\mathcal{P}_{N(\varepsilon)}\right)$$
(I left out maaany details. I hope it is still clear enough.)
Thus it is also Riemann integrable $F\in\mathcal{L}_\mathfrak{R}$.
Moreover, one has an upper bound:
$$\int\|F\|\mathrm{d}\mu\leq\|F\|_\infty\mu(\Omega)<\infty$$
Hence it is also Bochner integrable $F\in\mathcal{L}_\mathfrak{R}$.
Besides, the preceding construction gives:
$$\|\int_\mathfrak{R}F\mathrm{d}\mu-\int S_n\mathrm{d}\mu\|\leq\ldots\leq\|\int_\mathfrak{R}F\mathrm{d}\mu-\sum_{\mathcal{P}(\varepsilon)}F(a)\mu(A)\|\\+\|F-S_n\|_{\Omega_N}\mu(\Omega)+3\|F\|_\infty\mu(\Omega_N^\complement)<\varepsilon\quad(n\geq N(\varepsilon))$$
and therefore the integrals coincide:
$$\int_\mathfrak{R}F\mathrm{d}\mu=\lim_n\int S_n\mathrm{d}\mu=\int_\mathfrak{B}F\mathrm{d}\mu$$
Remark
Note that the above shows for finite measures:
$$\|F\|_\infty<\infty:\quad \|F-S_n\|\downarrow0\implies\int S_n\mathrm{d}\mu\to\int_\mathfrak{R}F\mathrm{d}\mu\quad(F\in\mathcal{L}_\mathfrak{R})$$
Caution that it may be not uniform limit:
$$F:(0,1]\to\mathcal{H}:F(\frac{1}{n+1}<t\leq\frac{1}{n}):=e_n$$
