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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.)

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  • Example: $f:(0,1]\to\mathbb{R}:f(\frac{1}{n+1}<x\leq\frac{1}{n}):=q_n\quad(q_n\in\mathbb{Q}\cap(0,1])$ – C-star-W-star Nov 12 '14 at 01:00
  • If you are taking the uniform limit (and not the $L^\infty (\mu)$-limit) in the first case, then every function integrable w.r.t. the first definition is bounded, whereas for the second definition, we only need $f \in L^\infty (\mu)$. Is this what you intended? – PhoemueX Nov 12 '14 at 20:30
  • @PhoemueX: No I intended the almost everywhere uniform limit ($\mathcal{L}^\infty$) but I kept it short so it won't get messy (I hoped it will be clear from the context). Will add a note now! – C-star-W-star Nov 13 '14 at 12:23
  • @PhoemueX: Got it! :D (See below.) – C-star-W-star Nov 21 '14 at 17:14

1 Answers1

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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

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