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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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  • $\begingroup$ Example: $f:(0,1]\to\mathbb{R}:f(\frac{1}{n+1}<x\leq\frac{1}{n}):=q_n\quad(q_n\in\mathbb{Q}\cap(0,1])$ $\endgroup$ – C-Star-W-Star Nov 12 '14 at 1:00
  • $\begingroup$ 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? $\endgroup$ – PhoemueX Nov 12 '14 at 20:30
  • $\begingroup$ @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! $\endgroup$ – C-Star-W-Star Nov 13 '14 at 12:23
  • $\begingroup$ @PhoemueX: Got it! :D (See below.) $\endgroup$ – C-Star-W-Star Nov 21 '14 at 17:14
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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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  • $\begingroup$ I don't have much time right now, but: Every bounded measurable function is the uniform limit of simple functions (if we allow "steps" that are arbitrary measurable sets, not just intervals (which would give Riemann step functions)). This would indicate that your counterexample is not a counterexample. $\endgroup$ – PhoemueX Nov 21 '14 at 17:16
  • $\begingroup$ Hmmm, right ... $\endgroup$ – C-Star-W-Star Nov 21 '14 at 17:20
  • $\begingroup$ @PhoemueX: I think I got it now. Can you check quickly? $\endgroup$ – C-Star-W-Star Nov 25 '14 at 1:25
  • $\begingroup$ I really don't understand the downvote. Does somebody have an idea what could be wrong? $\endgroup$ – C-Star-W-Star Nov 25 '14 at 18:36
  • $\begingroup$ I only looked over everything quickly, but it seems good to me especially the first part (especially that the Riemann integral vanishes) is interesting. $\endgroup$ – PhoemueX Nov 25 '14 at 19:13

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