I just rethought that...

.so continuing that thought $\delta A$ would be the length of all the bad rectangles, and $\delta^2 A^2$ would be the area of all the bad rectangles. So how would I end up with $\frac{2A\delta}{\sqrt{n}}$ as the bounding volume?

The sqrt is wrong. See errata. And remember that $A$ is the hyper-area. Think about dimensions.

Oh...I took the statement in errata to only be applying to the numerator

oh, so when i mess up it's a mistake, points off, bad grades, buzz off leslie nobody wants to be your friend anymore, but when ted is wrong, it's "errata"

You had the chance to bask in the glory of textbooks and theorems being created in your name, but you chose to abandon The Art and go into The Practice....

The advantage of publishing mistakes.

Well...that was painful.............but necessary.........

Well you may find amusing how I finally made headway in the problem..............I actaully gave up on the question, looked at #11 and #12 and gave up on those too and proceeded to do #14 on volume zero vs measure zero. I got to the second part asking about an example of a set (which I'm still on) and asked myself "what does a volume zero set look like?"....

then I went looking through the previous section in the text and began reading the proof of integrability over a rectangle which has a volume zero set. And lo and behold, the idea of "good" and "bad" rectangles emerged. Then looked at the proof of continuous implies integrable again and the idea of diameter $< \delta$......then I went and looked at the notes I took of that particular lecture and rewatched that portion of the lecture.....

bounds and estimates brain is at least stage 3 of one of those galaxy brain memes

category brain is the last stage

It is easily confused with dementia,

if you've gotta be on stage 3 of something, bounds and estimates brain is the one you want

maybe LHF