Simple question about Upper and Lower Riemann Integrals [duplicate]

First of all, here are Definitions 6.1, 6.2, and 6.3 in Baby Rudin, 3rd edition:

Definition 6.1:

Let $[a, b]$ be a given interval. By a partition $P$ of $[a, b]$ we mean a finite set of points $x_0, x_1, \ldots, x_n$, where $$ a = x_0 \leq x_1 \leq \cdots \leq x_{n-1} \leq x_n = b.$$ We write $$ \Delta x_i = x_i - x_{i-1} \qquad (i = 1, \ldots, n). $$ Now suppose $f$ is a bounded real function defined on $[a, b]$. Corresponding to each partition $P$ of $[a, b]$ we put $$ \begin{align} M_i &= \sup f(x) \qquad (x_{i-1} \leq x \leq x_i), \\ m_i &= \inf f(x) \qquad (x_{i-1} \leq x \leq x_i), \\ U(P, f) &= \sum_{i=1}^n M_i \Delta x_i, \\ L(P, f) &= \sum_{i=1}^n m_i \Delta x_i, \end{align} $$ and finally $$ \begin{align} \tag{1} \overline{\int}_a^b f dx &= \inf U(P, f), \\ \tag{2} \underline{\int}_a^b f dx &= \sup L(P, f), \end{align} $$ where the $\inf$ and the $\sup$ are taken over all partitions $P$ of $[a, b]$. The left members of (1) and (2) are called the upper and lower Riemann integrals of $f$ over $[a, b]$, respectively.

If the upper and lower integrals are equal, we say that $f$ is Riemann-integrable on $[a, b]$, we write $f \in \mathscr{R}$ (that is, $\mathscr{R}$ denotes the set of Riemann-integrable functions), and we denote the common value of (1) and (2) by $$ \tag{3} \int_a^b f dx, $$ or by $$ \tag{4} \int_a^b f(x) dx. $$ This is the Riemann integral of $f$ over $[a, b]$. Since $f$ is bounded, there exist two numbers, $m$ and $M$, such that $$ m \leq f(x) \leq M \qquad (a \leq x \leq b). $$ Hence, for every $P$, $$ m(b-a) \leq L(P, f) \leq U(P, f) \leq M (b-a), $$ so that the numbers $L(P, f)$ and $U(P, f)$ form a bounded set. This shows that the upper and lower integrals are defined for every bounded function $f$. . . .

Definition 6.2:

Let $\alpha$ be a monotonically increasing function on $[a, b]$ (since $\alpha(a)$ and $\alpha(b)$ are finite, it follows that $\alpha$ is bounded on $[a, b]$). Corresponding to each partition $P$ of $[a, b]$, we write $$ \Delta \alpha_i = \alpha \left( x_i \right) - \alpha \left( x_{i-1} \right). $$ It is clear that $\Delta \alpha_i \geq 0$. For any real function $f$ which is bounded on $[a, b]$ we put $$ \begin{align} U(P, f, \alpha) &= \sum_{i=1}^n M_i \Delta \alpha_i, \\ L(P, f, \alpha) &= \sum_{i=1}^n m_i \Delta \alpha_i, \end{align} $$ where $M_i$, $m_i$ have the same meaning as in Definition 6.1, and we define $$ \begin{align} \tag{5} \overline{\int}_a^b f d \alpha = \inf U(P, f, \alpha), \\ \tag{6} \underline{\int}_a^b f d \alpha = \sup L(P, f, \alpha), \end{align} $$ the $\inf$ and $\sup$ again being taken over all partitions. If the left members of (5) and (6) are equal, we denote their common value by $$ \tag{7} \int_a^b f d \alpha $$ or sometimes by $$ \tag{8} \int_a^b f(x) d \alpha(x). $$ This is the Riemann-Stieltjes integral (or simply the Stieltjes integral) of $f$ with respect to $\alpha$, over $[a, b]$.

If (7) exists, i.e., if (5) and (6) are equal, we say that $f$ is integrable with respect to $\alpha$, in the Riemann sense, and write $f \in \mathscr{R}(\alpha)$.

By taking $\alpha(x) = x$, the Riemann integral is seen to be a special case of the Riemann-Stieltjes integral. . . .

Definition 6.3:

We say that the partition $P^*$ is a refinement of [a partition] $P$ if $P^* \supset P$ (that is, if every point of $P$ is also a point of $P^*$). Given two partitions $P_1$ and $P_2$, we say that $P^*$ is their common refinement if $P^* = P_1 \cup P_2$.

Now using this machinery how can we evaluate the integral $$ \int_0^1 x^2 \ \mathrm{d} x? $$

Next, here are Theorems 6.4, 6.8, and 6.9:

Theorem 6.4:

If $P^*$ is a refinement of $P$, then $$ \tag{9} L(P, f, \alpha) \leq L \left( P^*, f, \alpha \right) $$ and $$ \tag{10} U \left( P^*, f, \alpha \right) \leq U( P, f, \alpha). $$

And, so we have $$ L(P, f, \alpha) \leq L \left( P^*, f, \alpha \right) \leq U \left( P^*, f, \alpha \right) \leq U( P, f, \alpha). $$

Theorem 6.6:

$f \in \mathscr{R}(\alpha)$ on $[a, b]$ if and only if for every $\varepsilon > 0$ there exists a partition $P$ such that $$ U(P, f, \alpha ) - L( P, f, \alpha ) < \varepsilon. $$

Theorem 6.8:

If $f$ is continuous on $[a, b]$, then $f \in \mathscr{R}(\alpha)$ on $[a, b]$.

Theorem 6.9:

If $f$ is monotonic on $[a, b]$ and $\alpha$ is continuous on $[a, b]$, then $f \in \mathscr{R}(\alpha)$. (We still assume, of course, that $\alpha$ is monotonic.)

The function $f(x) = x^2$ is of course continuous as well as monotonic on the interval $[0, 1]$. Thus by either Theorem 6.8 or Theorem 6.9, our integral exists of course.

My Attempt:

Let $$P = \left\{ x_0, x_1, \ldots, x_{n-1}, x_n \right\}, $$ where $$ 0 = x_0 < x_1 < \ldots < x_{n-1} < x_n, $$ be a partition of $[0, 1]$. Then as our function $f$ is sttictly increasing on $[0, 1]$, so we find that, for each $i = 1, \ldots, n$, we have $$ m_i = f \left( x_{i-1} \right) = x_{i-1}^2 \qquad \mbox{ and } \qquad M_i = f \left( x_i \right) = x_i^2. $$ [Refer to Definition 6.1 above for notation.] Therefore $$ L(P, f) = \sum_{i=1}^n x_{i-1}^2 \left( x_i - x_{i-1} \right) \qquad \mbox{ and } \qquad U(P, f) = \sum_{i=1}^n x_{i}^2 \left( x_i - x_{i-1} \right). $$

Now from these two formulas, can we compute the quantities in (1) and (2) in Definition 6.1 above? I have no idea of how we can.

However, we can do the following trick:

Let us put $$ h \colon= \min \left\{ \ \Delta x_1, \ldots, \Delta x_n \ \right\}. $$ Then of course this $h$ satisfies $$ 0 < h \leq 1, $$ from which we obtain $$ \frac{1}{h} \geq 1. $$ Now let us put $$ k = \left\lfloor \frac{1}{h} \right\rfloor + 1. $$ This $k$ is of course a natural number, and we also have the inequality $$ k-1 \leq \frac{1}{h} < k.$$ Now let $P^\prime$ be the partition of $[0, 1]$ given by $$ P^\prime \colon= \left\{ \ 0, \frac{1}{k}, \ldots, \frac{k-1}{k}, 1 \ \right\}, $$ and let $$ P^* \colon= P \cup P^\prime. $$ Then by Theorem 6.4 in Baby Rudin, we have the following two sets of inequalities: $$ L(P, f, \alpha) \leq L \left( P^*, f, \alpha \right) \leq U \left( P^*, f, \alpha \right) \leq U( P, f, \alpha). $$ And, $$ L \left( P^\prime, f, \alpha \right) \leq L \left( P^*, f, \alpha \right) \leq U \left( P^*, f, \alpha \right) \leq U\left( P^\prime, f, \alpha \right). $$

Now for the partition $P^\prime$, we compute $$ L \left( P^\prime, f, \alpha \right) = \frac{1}{k} \sum_{i=0 }^{n-1} \left( \frac{i}{k} \right)^2 = \frac{1}{k^3} \sum_{i=1}^{n-1} i^2 = \frac{ (k-1) (2k-1 ) }{6k^2} = \frac{1}{6} \left( 1 - \frac{1}{k} \right) \left( 2 - \frac{1}{k} \right), $$ and $$ U \left( P^\prime, f, \alpha \right) = \frac{1}{k} \sum_{i=1 }^n \left( \frac{i}{k} \right)^2 = \frac{1}{k^3} \sum_{i=1}^{n} i^2 = \frac{ (k+1) (2k + 1 ) }{6k^2} = \frac{1}{6} \left( 1 + \frac{1}{k} \right) \left( 2 + \frac{1}{k} \right). $$ And, the supremum of all the lower sums $L \left( P^\prime, f, \alpha \right)$ and the infimum of all the upper sums $U \left( P^\prime, f, \alpha \right)$ obtained in this manner each equals $1/3$.

How to prove from here (or using some other device) that $$ \int_0^1 x^2 \ \mathrm{d} x = \frac{1}{3}?$$

Basically what you have shown in the last block of text is that for the partitions $$P_n=\left(0,\frac{1}{n},\ldots,\frac{n-1}{n},1\right)\qquad (n\in\mathbb N)$$ we have $$U(P_n,f)=\frac{1}{6} \left( 1 + \frac{1}{n} \right) \left( 2 + \frac{1}{n} \right), \qquad L(P_n,f)=\frac{1}{6} \left( 1 - \frac{1}{n} \right) \left( 2 - \frac{1}{n} \right).$$ Since $$\lim_{n\to\infty}U(P_n,f)=\frac{1}{3}=\lim_{n\to\infty}L(P_n,f)$$ we have $\inf_P U(P, f)\leq\frac{1}{3}$ and $\sup_PL(P,f)\geq \frac{1}{3}$, but as you noted, $f$ is continuous, so the integral exists and $\inf_P U(P, f)=\sup_PL(P,f)$. Combining this, we have $$\frac{1}{3}\leq\sup_P L(P, f)=\inf_P U(P, f)\leq\frac{1}{3},$$ so equality holds throughout, and thus the integral equals $\frac{1}{3}$.

The integral should be $$\int_0^1 x^2 \,dx = \frac13$$

You don't have to look at all the partitions.

For $n \in \mathbb{N}$ define the partition $P_n$ with $x_i = \frac{i}n$ for $i = 0, 1, \ldots, n$.

We have $$m_i = \min f([x_{i-1},x_i]) = f(x_{i-1}) = \frac{(i-1)^2}{n^2}$$ $$M_i = \max f([x_{i-1},x_i]) = f(x_{i}) = \frac{i^2}{n^2}$$

so $$L(P_n, f) = \sum_{i=1}^n m_i\underbrace{(x_i - x_{i-1})}_{=\frac1n} = \frac1{n^3} \sum_{i=1}^n (i-1)^2 = \frac{(n-1)n(2n-1)}{6n^3} \xrightarrow{n\to\infty} \frac13$$ $$U(P_n, f) = \sum_{i=1}^n M_i\underbrace{(x_i - x_{i-1})}_{=\frac1n} = \frac1{n^3} \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6n^3} \xrightarrow{n\to\infty} \frac13$$

We conclude

$$\overline{\int_0^1} x^2 \,dx = \inf_Q U(Q, f) \le \frac13 \le \sup_Q L(Q, f) = \underline{\int_0^1} x^2\,dx$$

In general we know that $\underline{\int_0^1} x^2 \,dx \le \overline{\int_0^1} x^2\,dx$ so they are actually equal. It follows that the integral exists and is equal to $\frac13$.