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.
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.
