True /False : If $P^*$ is a refinement of $P$, then $U(P^* ,f,a) \le U (P,f,a)$

Question

I have confusion in Baby Rudin theorem $6.4$:

Let $P$ be a partition of $[a,b]$, that is $a=x_0 \le x_1 \le \ldots \le x_n = b$, $f$ is the function that we want to integrate, $\alpha$ is an increasing function on $[a,b]$, $M_i = \sup_{x_{i-1}\le x \le x_i} f(x)$ and $\Delta \alpha_i = \alpha(x_i) - \alpha(x_{i-1})$.

Define $$U(P, f, \alpha) = \sum_{i=1}^n M_i \Delta \alpha_i$$

Theorem: If $P^*$ is a refinement of $P$, then $U(P^* ,f,\alpha) \le U (P,f,\alpha)$

I think this statement is N0T true

My attempt : Let $f(x)= x$ .Take $P=\{0,\frac{1}{4} ,1\} $ and $P^*=\{0,\frac{1}{4} ,\frac{1}{2} ,1\} $

$w_1 = \sup \{f(x) | x \in [0,\frac{1}{2}]\}=\frac{1}{2}$

$w_2 = \sup \{f(x) | x \in [\frac{1}{2},1]\}=1$

$M_i = \sup \{f(x) | x \in [0,1]\}=1$

Then $U(P^*,f)- U(P,f)$ = $w_1 + w_2 - M_i =\frac{1}{2} +1 -1 \ge 0 \implies U(P^*,f)\ge U(P,f)$

Answer 1

$$w_1 = \sup \{f(x) | x \in [\frac14,\frac{1}{2}]\}=\frac{1}{2}$$

$$w_2 = \sup \{f(x) | x \in [\frac{1}{2},1]\}=1$$

$$M_i = \sup \{f(x) | x \in [0,1]\}=1$$

Then if $\alpha$ is an increasing function,\begin{align}&U(P^*,f)- U(P,f) \\ &= w_1 (\alpha(1) - \alpha(0.5)) + w_2(\alpha(0.5) - \alpha(0.25)) - M_i(\alpha(1)-\alpha(0.25)) \\&=\frac{1}{2}(\alpha(1) - \alpha(0.5)) +1(\alpha(0.5) - \alpha(0.25)) -(\alpha(1)-\alpha(0.25)) \\ &=\frac{1}{2}(\alpha(1) - \alpha(0.5)) +1(\alpha(0.5) - \alpha(0.25)) -(\alpha(1)-\alpha(0.5) + \alpha(0.5) - \alpha(0.25)) \\ &=-\frac12(\alpha(1) - \alpha(0.5)) \le 0 \end{align}

I think your fallacy is to assume that we can have

$$\alpha(1)-\alpha(0.25) = \alpha(1)-\alpha(0.5)=\alpha(0.5)-\alpha(0.25)=1$$

Answer 2

I don’t understand your computation. Why don’t you compute directly

$$\frac{11}{16}=\frac{1}{4}\frac{1}{4} + \frac{1}{4}\frac{1}{2} + \frac{1}{2}1=U(P^* ,f,a) \le U (P,f,a)= \frac{1}{4} \frac{1}{4} + \frac{3}{4}1 =\frac{13}{16} $$

By the way, I don’t have Rudin’s proof at hand of the theorem, but it is rather clear and simple.