# Reinventing The Wheel - Part 1: The Riemann Integral [closed]

Preface

The core of any notion of integral is some sort of weighted sum: $$\sum b\mu(A)$$

Depending on wether the domain or range is decomposed these split into Riemann and Lebesgue type ones: $$\{A_i\}_{I \in I}: b_i\in fA_i$$ $$\{b_i\}_{I \in I}:A_i=f^{-1}\{b_i\}$$

In order to handle kinds of infinite sums one can introduce a notion of convergence: $$\lim_{\leq}$$ The crucial ingredient for uniqueness is a directed order together with a Hausdorff space: $$\leq\text{ directed}\\ \mathcal{T}\text{ Hausdorff}$$ (For more details see Hausdorffness vs Uniqueness.)

(Besides, these two first facts, namely summation and convergence, are the ones that weak notions of integral exploit.)

For Lebesgue type ones the natural notion of convergence fails; that is plain summation gives: $$f:[0,1]\to\mathbb{R}:f(x):=x:\quad\{\sum_{b\in B}b\mu(f^{-1}\{b\})\}_{\# B_0<\infty}=0\qquad B_0\leq B_0':\iff B_0\subseteq B_0'$$ For Riemann type ones this subtlety does not arise since it foots on collections of sets rather than elements. That will be the starting point of this article!

Strategy & Goal

• Hausdorff topological vector space: $V$
• Measure space of possible infinite size: $\mu(\Omega)\leq\infty$
• Integral of Riemann type: $I(f)=\lim_{\leq}\sum_{A\in\mathcal{A}}f(b)\mu(A)$

The primary goal will be to entail functions with poles and oscillations while retaining uniformly continuous functionsof the following form:

• Uniformly continuous: $I(f)\text{ exists}$
• Poles: $f:(0,1]\to\mathbb{R}: f(x):=x^{\alpha>-1}$
• Oscillation: $g:[1,\infty)\to\mathbb{R}: g(x):=\frac{1}{x}\sin(x)$

Definition

Consider a sigma-finite complete measure space and a Banach space as well as Banach space valued functions: $$(\Omega,\Sigma,\mu)\text{ and }(E,\|\cdot\|) \text{ and }(\mathcal{F},\ldots)$$

Define the integral if it exists by: $$\int f\mathrm{d}\mu:=\lim_{(\Sigma_0,\leq)}\left\{\sum_{(A,a)\in\Sigma_0}\mu(A)f(a)\right\}_{\Sigma_0}$$ with the finite collections of disjoint measurable subsets of finite size: $$\Sigma_0\subseteq\Sigma: \qquad \#\Sigma_0<\infty\text{ as well as }\mu(A)<\infty\text{ and }A\cap B=\varnothing\text{ for }A\neq B,A,B\in\Sigma_0$$ being ordered by refinement and expansion: $$\Sigma_0\leq\Sigma_0':\iff\Sigma_0\dashv\Sigma_0',\Sigma_0\prec\Sigma_0'$$ where refinement and expansion are defined as: $$\Sigma_0\dashv\Sigma_0':\iff\forall A\in\Sigma_0\exists A'\in\Sigma_0':A\supseteq A'$$ $$\Sigma_0\prec\Sigma_0':\iff\cup_{A\in\Sigma_0}A\subseteq\cup_{A'\in\Sigma_0'}A'$$

In fact, these are considered as tagged collections but for better reading this is masked: $(A,a)\in\Sigma_0$
Note also that the collections are not required to actually cover the measure space: $\cup_{A\in\Sigma_0}A\subseteq\Omega$
The ordering of collections is chosen so to give the right result: $\int s\mathrm{d}\mu=\sum_{e\in \mathrm{im}s}\mu(s^{-1}\{e\})e$

Interpretation:

This can be interpreted as the net of simple functions: $$s_{\Sigma_0}:=\sum_{(A,a)\in\Sigma_0}f(a)\chi_A$$

Then the above read: $$\int s_{\Sigma_0}\mathrm{d}\mu\to\int f\mathrm{d}\mu$$

Moreover for integrable functions the net of simple functions converges pointwise (proof?): $$\left(\int s_{\Sigma_0}\mathrm{d}\mu \to \int f\mathrm{d}\mu\right) \Rightarrow \left(s_{\Sigma_0}(\omega)\to f(\omega)\right)$$ (Note that it is claimed pointwise convergence everywhere rather than almost everywhere.)

So the question arises how this Riemann type integral relates to the Lebesgue type integral by Bochner.
(The crucial difference is that this approach considers the net of simple functions related to the function under consideration while the approach by Bochner considers some sequence of simple functions not necessarily related to the function under consideration.)

Discussion:

It turns out that this notion of integrability is very restrictive.

A necessary condition on integrability is: $$\exists\mu(N)=0: \|f(\Omega\setminus N)\|<\infty$$

So though the first critical example won't be dissolved by this notion: $$f:(0,1]\to\mathbb{R}:x\mapsto \frac{1}{\sqrt{x}}$$ the second will be dissolved however: $$g:[1,\infty)\to\mathbb{R}:x\mapsto\frac{1}{x}\sin(x)$$

Especially no Lebesgue type integrability does imply this Riemann type integrability.

However, a positive result is achieved on finite measure spaces.

A sufficient condition on integrability is: $$\ldots$$

Absolute integrability does not imply integrability (example?).

Regarding convergence theorems the uniform convergence theorem holds (proof?) but the dominated, monotone and Fatou convergence theorem fail (example?).

Summary:

$$\ldots$$

• It would make more sense if you write ‘$\displaystyle I(s) \stackrel{\text{def}}{=} \sum_{e \in E} \mu({s^{-1}}[\{ e \}]) \cdot e$’. – Leonard Huang May 9 '14 at 20:37
• If you wish to develop Banach space-valued integration theory, you might want to use strongly measurable functions instead of just measurable functions in the sense of Rudin. This is because a Banach space may not be separable. – Leonard Huang May 9 '14 at 20:45
• The values are elements of the Banach space (that is sloppy terminology usually used) – C-Star-W-Star May 12 '14 at 19:39
• I don't want to be raining on your parade, but this seems more like something you should be posting on a blog. I don't see any question or whatever, just some exposition. – Pedro Tamaroff Aug 3 '14 at 21:53
• @PedroTamaroff: Yeah I know - it started as a question and become a dead end as it failed at a critical point. Will vote to close it now... – C-Star-W-Star Aug 3 '14 at 22:02

Attempt:

Assume every point admits a subset of finite measure (sigma-finite measure spaces do so) that is every point comes into consideration eventually: $\forall\omega\in\Omega\exists\Sigma_0: \omega\in A\in\Sigma_0$

Fix the point and consider two cases.

All measurable subsets containing the point are of bounded size: $\mu(A_{\omega})\geq C>0$
Let $\epsilon>0$. Choose $\delta=C\epsilon$. Find $\Sigma_\delta$. Assume w.l.o.g. $\omega\in A\in\Sigma_\delta$. Regard $\Sigma_a\geq\Sigma_\delta$. Retag $\Sigma_\omega=\Sigma_a$.
So it follows: $$\|f(a)-f(\omega)\|<\frac{\delta}{\mu(A_\omega)}\leq\frac{C\epsilon}{C}=\epsilon$$

Some measurable subsets containing the point tend to zero size: $\mu(A_n)\to 0$
Since the intersection has zero measure the singleton is measurable by completeness with finite size: $\mu(\omega)<\infty$
Thus it holds: $$s_{\Sigma_0}(\omega)=f(\omega)\text{ for all }\Sigma_0\geq\{(\{\omega\},\omega)\}$$

Actually the ordering gives: $$f:[0,1]\to\mathbb{R}:f(x):=x:\quad\text{ not integrable}$$