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In the book of Mathematical Analysis 2 by Zorich, at page 117, it is given that

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However, I couldn't understand the existence of the number $\lambda_{\epsilon}$ and in general what the author is trying to do with $\Gamma_\epsilon$. Can someone explain it to me ? or give some hints ?

Edit:

For additional questions, please see my comments on the given answer to the question.

Our
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  • The proof is unnecessarily difficult and uses sets of measure zero. One can provide a direct proof as given in this answer : https://math.stackexchange.com/a/2047959/72031 – Paramanand Singh Feb 03 '18 at 09:49
  • @ParamanandSingh Thanks a lot really. I mean I have been dealing the proof of this theorem since I have started studying this boo :) – Our Feb 03 '18 at 10:07

1 Answers1

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The point is that for a partition $P$ with $\lambda(P)$ small enough, the lower Darboux sum $s(f,P)$ cannot be much smaller than $s(f,P_{\varepsilon})$. We can distinguish the partition intervals of $P$ according to whether they are completely contained in a partition interval of $P_{\varepsilon}$ or not. Clearly if $J_1 \subset J_2$, then

$$\inf\: \{ f(x) : x \in J_1\} \geqslant \inf\: \{ f(x) : x \in J_2\}\,,$$

so the part of the lower Darboux sum $s(f,P)$ corresponding to intervals contained in a $P_{\varepsilon}$-interval is not smaller than the corresponding part of $s(f,P_{\varepsilon})$. Thus "bad things" can only occur on intervals containing a partition point of $P_{\varepsilon}$. But we can control how bad the things that might happen there are. For every subset $S$ of $I$, we have $\inf\:\{ f(x) : x \in I\} \leqslant \inf\:\{ f(x) x \in S\} \leqslant \sup\:\{ f(x) : x \in I\}$, and therefore

\begin{align} \lvert \inf\:\{ f(x) : x \in J\} - \inf\:\{ f(x) : x \in J'\}\rvert &\leqslant \sup\:\{ f(x) : x \in I\} - \inf\:\{ f(x) : x \in I\} \\ &\leqslant 2\,\underbrace{\sup\:\{\lvert f(x)\rvert : x \in I\}}_M\,. \end{align}

So the worst that can happen is that the contribution to $s(f,P)$ of an interval of $P$ that contains a partition point of $P_{\varepsilon}$ is $2M$ times the length of the interval smaller than the contribution of this part of $I$ to $s(f,P_{\varepsilon})$. That adds up to

$$s(f,P) \geqslant s(f,P_{\varepsilon}) - 2M\sum_J \ell(J)\,,$$

where the sum is over those partition intervals of $P$ that contain a partition point of $P_{\varepsilon}$, and $\ell(J)$ is the length of the interval $J$.

Now there are only finitely many partition points of $P_{\varepsilon}$, say the number is $k$. And every partition point of $P_{\varepsilon}$ belongs to at most two closed partition intervals of $P$, so we find

$$s(f,P) \geqslant s(f,P_{\varepsilon}) - 2M\cdot 2k\cdot \lambda(P)\,.$$

And it's obvious that by requiring $\lambda(P)$ to be small enough, we can make the lower bound as close to $s(f,P_{\varepsilon})$ as we please. For example, with $\lambda_{\varepsilon} = \frac{\varepsilon}{4Mk}$ we get

$$s(f,P) > s(f,P_{\varepsilon}) - \varepsilon$$

for $\lambda(P) < \lambda_{\varepsilon}$, whence

$$\liminf_{\lambda(P) \to 0}\: s(f,P) \geqslant s(f,P_{\varepsilon}) - \varepsilon \geqslant \underline{\mathcal{J}} - 2\varepsilon\,.$$

Daniel Fischer
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  • After reading your answer several times, I have some questions – Our Feb 02 '18 at 06:25
  • First of all, I couldn't understand what kind of "bad things" can occur at the partition point of $P$ ? – Our Feb 02 '18 at 06:26
  • Secondly, are we choosing $\lambda(P)$ small enough that even the "largest" interval of $P$ is contained in the "smallest" interval of $P_\epsilon$ ? Otherwise, I can't see how can "every partition point of $P_\epsilon$ belong to at most two closed partition intervals of $P$" ? I mean the fact that $s(f, P_\epsilon) \leq s(f, P)$ only implies that at least some of the intervals of $P$ are contained in some of the intervals of $P_\epsilon$, but this does not mean that other intervals of $P$ contain some intervals of $P_\epsilon$. – Our Feb 02 '18 at 06:45
  • Thirdly, why is the number of partition points of $P$ finite ? I mean the index of Riemann sum is over $\mathbb{N}$ , so the set of partition points of $P$ is at most countable. – Our Feb 02 '18 at 06:51
  • it would be nice if you answered to my comments, hence I could accepted your answer. – Our Feb 13 '18 at 07:03
  • Sorry for the delay, my laptop decided that it didn't want to connect to the internet any more at the end of January, so I wasn't online for a while. The "bad things" are that the step functions corresponding to the partition $P$ can be smaller (for the lower Darboux sum) or larger (for the upper sum) than the step functions corresponding to the partition $P_{\varepsilon}$. The difference of these functions is controlled by a bound for $\lvert f\rvert$, and the contribution to the Darboux sum by that and the length of the interval in question. So if we get the total length of such intervals – Daniel Fischer Mar 03 '18 at 14:54
  • small, we get the Darboux sums for $P_{\varepsilon}$ not much smaller/larger than those for $P$. Concerning the second point, the idea is to make the mesh of $P$ so small that each $P$-interval can contain at most one problematic point (that is, one point where "bad things" in the sense above can happen). But in view of your third question, that may require a modification. I was assuming that, as is standard, one considers partitions of the interval $I$ into finitely many parts. (There's a good reason why this is standard: It's simpler.) – Daniel Fischer Mar 03 '18 at 14:54
  • If Zorich uses a different approach and considers partitions with infinitely many points, could you tell me what definitions he uses? – Daniel Fischer Mar 03 '18 at 14:54