Talk:Fundamental Theorem of Calculus/Second Part/Proof 2

Thank you so much for fixing up the alternative second part proof! Hopefully I'll learn enough from your changes that my next proof will be better.

Question: when you say

"$\left[{a .. b}\right]$ can be divided into any number of closed subintervals of the form $\left[{x_{k-1} .. x_k}\right]$ where:


 * $ a = x_0 < x_1 \cdots < x_{k-1} < x_k = b$

Fix such a subdivision of the interval $\left[{a .. b}\right]$; call it $P$."

Is that the same thing as what my book is saying? The book I got the proof from, Larson/Hostetler/Edwards. It calls the $\Delta$ "[a] partition of $[a,b]$. I think $\Delta$ is another name for what the wiki proof calls $P$, am I right? Also, instead of talking about supremum and infimum, it takes the limit as $||\Delta||\to0$ and from what I understand from my book it means $||\Delta_{biggest}||\to0$. Are those the same thing?

--GFauxPas 10:42, 23 October 2011 (CDT)


 * A partition of $[a..b]$ (which is preferred over $[a,b]$ on PW) is indeed the same as a subdivision of $[a..b]$ (compare the definition). The supremum and infimum are not quite the same as $\|\Delta\| \to 0$. They are dependent on the particular chosen $P$, and are used in the definition of upper and lower sums. The trick I used to avoid that somewhat sloppy use of continuity is that I rewrote it into terms for which the definition of definite integral applies, i.e.:

"Suppose that $\exists y \in \R$ such that:

Then ... $\displaystyle y = \int_a^b f \left({x}\right)\ \mathrm d x$"
 * For any lower sum $L \left({P}\right)$ over any of subdivision $P$ of $\left[{a .., b}\right]$, $L \left({P}\right) \le y$
 * For any upper sum $U \left({P}\right)$ over any of subdivision $P$ of $\left[{a .. b}\right]$, $U \left({P}\right) \ge y$


 * which gives the conclusion immediately. It is a matter of style which definition of the integral one uses. The one used here at PW is very nice, as it avoids cluttering with convergence of some $||\Delta_{biggest}||$ to 0. It is however, hopefully, clear that these definitions are equivalent (some clutter on refinement of subdivisions and such), so I exploit the easier form for this occasion. That clear things up? --Lord_Farin 10:55, 23 October 2011 (CDT)

Haha not really! But I'm going to study Riemann sums and related subjects and then go back and reread your explanation :) I didn't learn about convergence in my class yet so I may go learn it on my own, depending on my ambition. Thanks though. --GFauxPas 13:58, 23 October 2011 (CDT)

I realize that I might have given the impression I'm writing out proofs straight from my book without understanding their content. That's not the case, and actually my presentation has been a bit different than the way my book does it, I just never learned about certain things and I need time to research it for myself. --GFauxPas 14:13, 23 October 2011 (CDT)