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In the proof of Theorem 5.5 on the below image, where does the n/6A term come from?

(click to see full size version)

enter image description here

The whole books might be download here:

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How did you beautify the picture, @Zev? – Tim Dec 28 '12 at 16:26
I googled the first sentence to figure out what book this was, then ... ahem ... scanned it myself and uploaded the picture. – Zev Chonoles Dec 28 '12 at 16:29
@ZevChonoles - Thank you so much! – Victor Dec 28 '12 at 16:33
Nice @Zev, although I am curious what "... ahem .." is. Victor, it seems like your question might be answered by reading Theorem 3.4 as pointed out in the proof. – Tim Dec 28 '12 at 16:34
@Tim The number 6 never appeared in Theorem 3.4 and its proof! – Victor Dec 28 '12 at 16:35
up vote 2 down vote accepted

Expanding on my comment, this is what I think it means. It would be useful to have Theorem 3.4 to be able to be completely sure, but I think the idea is that for any $\epsilon \gt 0$, $\exists \delta$ such that given any partition $P$ with $\text{mesh}\ P \lt \delta$, the difference between the Riemann sum associated with $P$ and the integral is less than $\epsilon$. Here, we take $\epsilon$ to be $\frac{\eta}{12A}$, and we know that $\exists \delta_2$ satisfying the above condition. Why $\frac{\eta}{12A}$? Without seeing the next page of the book, I can only guess that eventually that $12A$ will go away and we'll be left with $\eta$ alone. If you had started with $\eta$ instead of $\frac{\eta}{12A}$, we would end up with an extra $12A$ around.

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