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I have a doubt in one step of the proof of the following theorem in Rudin's Principles of Mathematical Analysis enter image description here

Rudin proceeds to prove this as follows:

Given $\epsilon>0, \exists$ a partition P such that $$U(P,\alpha')-L(P,\alpha') < \epsilon$$ Let $M=sup|f(x)|$

He then proves that $$ |U(P,f,\alpha)-U(P,f\alpha')| \leq M\epsilon$$

Also, he has shown that the above is true for any refinement of $P$ as well.

It is clear to me till here. But then he 'concludes' that

enter image description here

I fail to understand how this follows from the the fact that the inequality for $P$ is also true for any refinement of $P$.

Please help!

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Suppose

$$ \overline {\int_a^b}f d\alpha - \overline {\int_a^b}f\alpha' dx > M\epsilon. $$

Since the upper integral is the infimum of the corresponding upper sums, we can choose a partition $\hat{P}$ such that

$$ \overline {\int_a^b}f d\alpha - U(\hat{P}, f\alpha') > M\epsilon. $$

Let $P$ be the partition from the proof. Then

$$ U(P, f, \alpha) - U(\hat{P}, f\alpha') \geq \overline {\int_a^b}f d\alpha - U(\hat{P}, f\alpha') > M\epsilon. $$ Now $P \cup \hat{P}$ is a refinement of $P$, and also

$$ U(P \cup \hat{P}, f, \alpha) - U(\hat{P}, f\alpha') \geq \overline {\int_a^b}f d\alpha - U(\hat{P}, f\alpha') > M\epsilon. $$

Since

$$ U(P \cup \hat{P}, f\alpha') \leq U(\hat{P}, f\alpha'), $$

we now have that

$$ U(P \cup \hat{P}, f,\alpha) - U(P \cup \hat{P}, f\alpha') \geq U(P \cup \hat{P}, f, \alpha) - U(\hat{P}, f\alpha') > M\epsilon, $$

which is a contradiction to the earlier conclusion in the proof.

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  • $\begingroup$ Hi, I'm sorry but how does this contradict the fact that $|U(P,f,\alpha) - U(P,f\alpha')|\leq M\epsilon$? $\endgroup$ – Thomas Winckelman Aug 24 at 12:25
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The purpose of proving $\left| \overline\int_a^b fd\alpha - \overline\int_a^b f(x)\alpha'(x)dx \right|\le M\epsilon$ is to show $\overline\int_a^b fd\alpha = \overline\int_a^b f(x)\alpha'(x)dx$. To this end, an alternative straightforward proof is as follows: For any $\delta > 0$, clearly we can find $P_1, P_2$ such that $\left| \overline\int_a^b fd\alpha - U(P_1, f, \alpha) \right|\le \delta$ and $\left| \overline\int_a^b f(x)\alpha'(x)dx - U(P_2, f\alpha') \right|\le \delta.$ Recall also that $\left| U(P, f, \alpha) - U(P, f\alpha') \right|\le M\epsilon$. Now let $P^*=P\cup P_1 \cup P_2$. Then these three inequalities still hold, if we substitute $P^*$ for $P_1, P_2,$ and $P$. So it follows that $$\left| \overline\int_a^b fd\alpha - \overline\int_a^b f(x)\alpha'(x)dx \right|\le M\epsilon+2\delta,$$ which shows $\overline\int_a^b fd\alpha = \overline\int_a^b f(x)\alpha'(x)dx$, since both $\epsilon$ and $\delta$ can be made arbitrarily small.

Remark: The last inequality also implies $\left| \overline\int_a^b fd\alpha - \overline\int_a^b f(x)\alpha'(x)dx \right|\le M\epsilon,$ since $\delta$ is arbitrary.

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  • $\begingroup$ Greetings, I like this a lot. If I could have anyone's attention, though, the one thing I don't see is how $|U(P,f,\alpha) - U(P,g,\beta)| \leq \varepsilon \implies |U(P^*,f,\alpha) - U(P^*,g,\beta)| \leq \varepsilon$ for a refinement $P^*$ of $P$ (meaning $P^* \supseteq P$). Can anyone enlighten me on this? I've walked myself through the other steps and I'm stuck on that one. $\endgroup$ – Thomas Winckelman Aug 22 at 23:04
  • $\begingroup$ We certainly do not have $\big(a \leq a' \wedge b \leq b' \big) \implies \big| a-b \big| \leq \big| a'-b' \big|$ for arbitrary real numbers, as we would seemingly like to apply to $a=U(P^*,f,\alpha), \hspace{1mm} a' =U(P,f,\alpha)$, etc.; take $a'=b'=2$ and $a=1$, $b=0$ for a counterexample. Yet, what else are we working with besides the fact that $U(P^*,g,\beta) \leq U(P,g,\beta)$ and $U(P^*,f,\alpha) \leq U(P,f,\alpha)$, which is a property of refinements (if I'm not mistaken)? $\endgroup$ – Thomas Winckelman Aug 22 at 23:39

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