Continuing from my last question, I understand my definition is unclear, so my tutor helped modify. Forgive him if it's still unclear; he's the best I have.
I asked the same question on Math Stack Exchange in case someone there can answer.
Definition
Consider $f:A\to[0,1]$ where $A\subseteq[0,...
EDIT: Willie Wong gave an example showing a flaw in my original definitions.
There are fractal like constructions that allows very frequently half of the intervals to be empty, and not in a way you can merge. Let B denote all infinite subsets of the square numbers $\{1, 4, 9, ...\}$. Let $A:...
In the new version, the definition is needlessly complicated.
2
Given $A \subset [0,1]$, and let $P$ be a partition of $[0,1]$ (note: a partition is a finite set of disjoint intervals), you can define $P' = \{ X\in P: X\cap A \neq \emptyset\}$. And you can define $n' = |P'|$ (the cardinality of a finite set).
In the new question, you seem to no longer require that the infimum and supremum be taken on $X \cap A$, just $X$. Is that the intention?
(Or is that a mistake?)
Notice that the definition above does not require looking at all at the number $s$. Given a partition $P$ you can define its norm as $\|P\| = \sup_{X\in P} \|X\|$, where $\|X\|$ of an interval is its length. So you can ask about the limits taken as $\|P\| \to 0$.
The upper and lower sums are still not monotonous under refinements, due to the factor of $n'$, so the fact that $P$ forms a directed set under refinement does not help much with taking the limits.
@WillieWong I apologize, you are my only hope. Otherwise, I would surely be ignored. I am an undergraduate student so I am the lowest of the low. Thank you so much for helping!
@WillieWong When you say $n'=|P'|$ (the cardinality of a finite set), do you mean $n'$ gives a finite number of sub-intervals $X$ which are in $P'$?
$P$ is a finite set of size $n$ (in your case where you divide the interval into $n$ equal subintervals). So $P'$ is a set of size at most $n$. The size of $P'$ is $n'$.
Probably my final comment on this: I think you can probably achieve what you want by looking at the problem differently.
First we construct a sequence of bounded functions $g_\sigma$ as follows: start with your $A$. consider the set $A_\sigma = \cup_{x\in A} (x - \sigma, x+\sigma)$. This is a union of of open intervals and hence is an open set. As long as $A$ is non-empty this set is non-empty, and hence has positive Lebesgue measure.
Consider only $\sigma < 1/2$. Let $\chi_\sigma$ be the indicator function of $A_\sigma$. Define $g_{\sigma}(x) = \frac{1}{|A_\sigma|} \int_{-1/2}^x \chi_\sigma(y) ~dy $.
Here $|A_\sigma|$ is the Lebesgue measure of $A_\sigma$.
Notice that $g_\sigma$ is normalized so that it takes value between $0$ and $1$. (It is bounded.)
And $g_\sigma$ is continuous. The question is whether there is, and what is, the limit $\lim_{\sigma\to 0} g_{\sigma}$.
When $|A| > 0$, then the family $g_\sigma$ is equicontinuous, and it is not too hard to see that $g$ is formed as \frac{1}{|A|} \int_{-1/2}^x \chi(y) ~dy $ and here $\chi(y)$ is the indicator function of $A$.
The main question is what happens when $|A| = 0$. The conjecture is that when $A$ has measure zero, but has a non-trivial perfect kernel, then the limiting $g$ is a continuous function (like the Cantor function). And when $A$ is scattered, the limiting $g$ is a step function. In either case, the integral you are looking for should be the Stieltjes integral with weight function $g$.
This not the same as "Coming up with a rigorous definition for a riemman-like sum which is easier to compute?
". Here I'm assuming my riemman-like sum is clear enough to understand. If not, try and answer this question.
Consider $f:A\to[0,1]$ where $A\subseteq[a,b]$ and $\lambda$ is the Lebesg...
New Discovery: "Riemman-like" Sum
New Discovery: "Riemman-like" Sum