Doubt in Baby Rudin Theorem 6.17 (Riemann-Stieltjes integral)

Question

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

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

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!

Answer 1

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.

Answer 2

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.