# How to evaluate this Riemann integral using this definition?

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 $$\int_0^1 x^2 \ \mathrm{d} x = \frac{1}{3}?$$

• Isn't sup of all lower sums and the inf of all upper sums equal to $1/3$? May 31 '18 at 15:02
• @caverac absolutely. Thank you for suggesting the correction. I'll edit my post asap. May 31 '18 at 15:15
• You may have a look at this answer where it is proved that the infimum of all upper Darboux sums is equal to the limit of these sums as the norm of partition tends to $0$. Thus it is sufficient to take limit of a Darboux sum over uniform partition with $n$ sub-intervals as $n\to\infty$. This does not assume that the function is Riemann integrable. Jun 1 '18 at 3:47

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 $\frac{1}{3}$.

• this is exactly what I'm not able to show. So can you please expand your answer by including this detail? May 31 '18 at 15:18
• Just to be clear, you want me to explain why $\lim_{n\to\infty}U(P_n,f)=\frac{1}{3}=\lim_{n\to\infty}L(P_n,f)$ implies $\inf_P U(P, f)=\frac{1}{3}=\sup_PL(P,f)$? May 31 '18 at 15:18
• exactly. I'd be grateful! May 31 '18 at 16:26
• I believe I have done that since your last comment. Would you mind reading it again to see if it helps? May 31 '18 at 16:27
• thanks. Yes, got it! I've accepted your answer as well. May 31 '18 at 16:30

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 $\frac13$.

• That we don't need to consider all partitions is not obvious. The infimum of a set $A$ may be greater than the infimum of set $B$ if $A\subseteq B$. Jun 1 '18 at 3:50
• @ParamanandSingh I agree, the nontrivial part being that $\underline{\int_a^b}f \le \overline{\int_a^b}f$ in general. It suffices to prove that for any partitions $P, Q$ we have $L(P,f) \le U(Q,f)$. Consider the mutual refinement $P \cup Q$. We have $$L(P, f) \le L(P \cup Q, f) \le U(P \cup Q, f) \le U(Q, f)$$ because lower Darboux sums increase when refined, and upper Darboux sums decrease (the OP stated this in Theorem $6.4$). Jun 1 '18 at 6:48