New Discovery: "Riemman-like" Sum

Also, please spell my name correctly. It is rather disrespectful what you are doing.

In the new version, the definition is needlessly complicated.

I have mentioned them simply because, looking at your wording, it seems unlikely that you actually wanted to link to the same post twice.

BTW you wrote about Willie Wong in your question that "he is on MathOverflow". Actually, he used to be a moderator on Mathematics.

Sorry that I have comments to form rather than to the content. (Still, improving the form might help your questions to be better received.)

> Edit: I would also like to know why this and this are unclear.

When I looked at your post, you have there twice the link to the same question.

If you use an initial partition of size $1/n$ for $n$ large, the first part of the set will be covered by $\approx \sqrt{n}$ intervals, the second part of the set by $\approx \log(n)^2$ intervals. So in your weighted average the first part is weighed infinitely more than the second, even though it is a first order limit point.

Going to your original motivation in your first/second question, there seems to be an issue. Say you start with the following set: $\{ 1/k: k \geq 5\} \cup \{ 1/2 + 2^{-j} + 2^{-k} : j, k \geq 5\}$. Your earlier analysis would force $1/2$ as the only second order limit point. But I am pretty sure if you compute your upper/lower sums, you will find that the simple limit point at 0 will be preferred.

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 $g$.

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 $A$.

Notice that $g_\sigma$ is normalized so that it takes value between $0$ and $1$. (It is bounded.)

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 $.

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.

Probably my final comment on this: I think you can probably achieve what you want by looking at the problem differently.