Banach Spaces: Uniform Integral vs. Riemann Integral Problem
Given a finite measure space $\Omega$ and a Banach space $E$.

One has strict inclusion:
  $$\mathcal{L}_\mathfrak{U}(\mu)\subsetneq\mathcal{L}_\mathfrak{R}(\mu):\quad\int_\mathfrak{U}F\mathrm{d}\mu=\int_\mathfrak{R}F\mathrm{d}\mu$$
  How to prove this from scratch?

Uniform Integral
Predefine the simple integral:
$$S=\sum_kb_k\chi(A_k):\quad\int_\mathfrak{S}S\mathrm{d}\mu:=\sum_k b_k\mu(A_k)$$
It is uniformly bounded:
$$\|\int_\mathfrak{S}S\mathrm{d}\mu\|\leq\|S\|_\infty\mu(\Omega)$$
So define the uniform integral by:
$$F=\lim_nS_n:\quad\int_\mathfrak{U}F\mathrm{d}\mu:=\lim_n\int_\mathfrak{S}S_n\mathrm{d}\mu$$
(More precisely, by the a.e. uniform closure!)
Riemann Integral
Define the Riemann integral by:
$$\int_\mathfrak{R}F\mathrm{d}\mu:=\lim_\mathcal{P}\{\sum_{a\in A\in\mathcal{P}}F(a)\mu(A)\}_\mathcal{P}$$
Finite measurable partitions:
$$\mathcal{P}\subseteq\Sigma:\quad\Omega=\bigsqcup_{A\in\mathcal{P}}A\quad(\#\mathcal{P}<\infty)$$
Order them by refinement:
$$\mathcal{P}\leq\mathcal{P}':\iff\forall A'\in\mathcal{P}'\exists A\in\mathcal{P}:\quad A\supseteq A'$$
(That is the usual ordering.)
 A: Ok, I think I got it now...
Strictness
Consider the famous example:
$$F:(0,1]\to\ell^2(0,1]:t\mapsto\chi_t$$
Then it can't be uniform limit as:
$$\|\chi_s-\chi_t\|=\frac{1}{\sqrt{2}}\quad(s\neq t)$$
Choose the partition:
$$\mathcal{P}\geq\mathcal{P}_\varepsilon:=\left\{\left(\frac{k-1}{K(\varepsilon)},\frac{k}{K(\varepsilon)}\right]:k=1,\ldots,K(\varepsilon)\right\}$$
So it is Riemann integrable as:
$$\|\sum_{a\in A\in\mathcal{P}}\chi(a)\lambda(A)\|^2=\sum_{a\in A\in\mathcal{P}}\lambda(A)^2\leq\frac{1}{K(\varepsilon)}\lambda(0,1]<\varepsilon$$
(Besides, its Riemann integral vanishes.)
Inclusion
Consider a uniform limit:
$$S_n\in\mathcal{S}:\quad S_n\to F$$
Choose a simple function:
$$S_{N(\varepsilon)}=\sum_{k=1}^{K}b_k\chi(A_k):\quad\|F-S_{N(\varepsilon)}\|<\frac{\varepsilon}{2\mu(\Omega)}$$
Choose the partition: 
$$\mathcal{P}\geq\mathcal{P}_\varepsilon:=\{A_1,\ldots,A_{K}\}$$
Then for finer partitions:
$$\|\sum_{a\in A\in\mathcal{P}}F(a)\mu(A)-\int_\mathfrak{U}F\mathrm{d}\mu\|\\
\leq\|\sum_{a\in A\in\mathcal{P}}F(a)\mu(A)-\sum_{a\in A\in\mathcal{P}}S_N(a)\mu(A)\mathrm{d}\mu\|+\|\int_\mathfrak{S}S_{N}\mathrm{d}\mu-\int_\mathfrak{U}F\mathrm{d}\mu\|\\
\leq\|F-S_N\|_\infty\mu(\Omega)+\|F-S_N\|_\infty\mu(\Omega)<\varepsilon$$
Especially, the integrals coincide.
Supplementary
Another potential example is given in: Stone's Theorem Integral
