Riemann Integral

Definition (upper sum, lower sum)

Let $[a,b]$ with $a<b$ be a closed and bounded interval in $\mathbb{R}$. By a partition of $\mathscr{P}$ of $[a,b]$ we mean a finite set of points

\[\begin{align} \mathscr{P}=\{x_0, x_1,\ldots, x_n\} \text{ such that } a = x_0 < x_1 < \cdots < x_n =b. \end{align}\]

There is no requirement that the points $x_i$ be equally spaced. For each $i=1,2,\ldots,n$, set

\[\begin{align} \Delta x_i = x_i - x_{i-1} \end{align}\]

which is equal to the length of the interval $[x_{i-1}, x_i]$.

Suppose that $f$ is bounded real valued function on $[a,b]$. Given a partition

\[\begin{align} \mathscr{P}=\{x_0, x_1, \ldots x_n\} \end{align}\]

of $[a,b]$, for each $i=1,\ldots,n$, let

\[\begin{align} \begin{split} m_i &= \inf\{f(t): t\in [x_{i-1},x_i]\} \\ M_i &= \sup\{f(t):t\in [x_{i-1},x_i]\} \end{split} \end{align}\]

Since $f$ is bounded, by the least upper bound property, $m_i, M_i$ are real value. If $f$ is continuous function on $[a,b]$, then by Corollary 4.2.9 for each $i$ there exist point s $t_i, s_i \in [x_{i-1},x_i]$ such that $M_i = f(t_i), m_i = f(s_i)$.

The upper sum $\mathscr{U}(\mathscr{P},f)$ for the partition $\mathscr{P}$ and function $f$ is defined by

\[\begin{align} \mathscr{U}(\mathscr{P}, f) = \sum_{i=1}^n M_i \Delta x_i \end{align}\]

Similarly, the lower sum $\mathscr{L}(\mathscr{P},f)$ is defined by

\[\begin{align} \mathscr{L}(\mathscr{P}, f) = \sum_{i=1}^n m_i \Delta x_i \end{align}\]

Since $m_i \leq M_i$ for all $i=1,\ldots, n$, we always have

\[\begin{align} \mathscr{L}(\mathscr{P},f) \leq \mathscr{U}(\mathscr{P},f) \end{align}\]

Definition 6.1.1

Let $f$ be a bounded real-valued function on the closed and bounded interval $[a,b]$. The upper and lower integrals of $f$, denoted $\overline{\int_a^b}f$ and $\underline{\int_a^b}f$ respectively, are defined by

\[\begin{align} \begin{split} \overline{\int_a^b}f &=\inf\{\mathscr{U}(\mathscr{P},f): \mathscr{P}\text{ is a partition of } [a,b]\} \\ \underline{\int_a^b}f &=\sup\{\mathscr{L}(\mathscr{P},f):\mathscr{P}\text{ is a partition of } [a,b]\} \end{split} \end{align}\]

Definition 6.1.2

A partition of \(\mathscr{P}^{*}\) of \([a,b]\) is a refinement of \(\mathscr{P}\) if \(\mathscr{P}\subset \mathscr{P}^*\).

Remark

A refinement of a given partition $\mathscr{P}$ is obtained by adding additional points to $\mathscr{P}$. If $\mathscr{P}_1$ and $\mathscr{P}_2$ are two partitions of $[a,b]$, then $\mathscr{P}_1\cup\mathscr{P}_2$ is a refinement of both $\mathscr{P}_1$ and $\mathscr{P}_2$.

Lemma 6.1.3

If $\mathscr{P}^*$ is a refinement of $\mathscr{P}$, then

\[\begin{align} \mathscr{L}(\mathscr{P}, f) \leq \mathscr{L}(\mathscr{P}^*, f) \leq \mathscr{U}(\mathscr{P}^*,f )\leq \mathscr{U}(\mathscr{P},f) \end{align}\]

<proof> Suppose

\[\begin{align} \mathscr{P}=\{x_0, x_1, \ldots, x_n\}, \mathscr{P}^* = \mathscr{P}\cup \{x^*\} \end{align}\]

where \(x^*\neq x_j\) for any \(j=0,1,\ldots,n\). Then there exists an index \(k\) such that \(x_{k-1} < x^* <x_{k}\). Let

\[\begin{align} \begin{split} M^1_k &= \sup\{f(t):t\in [x_{k-1}, x_k]\}\\ M^2_k &= \sup\{f(t): t\in [x_{k-1},x_k]\} \end{split} \end{align}\]

Since $f(t) \leq M_k$ for all $t\in [x_{k-1},x_k]$, we have that $f(t)\leq M_k$ for all \(t\in[x_{k-1},x^*]\) and also for all \(t\in [x^*, x_k]\). Thus both \(M^1_k\) and \(M^2_k\) are less than or equal to \(M_k\). Now

\[\begin{align} \mathscr{U}(\mathscr{P},f)=\sum_{j=1}^{k-1}M_j \Delta x_j + M^1_k(x^*-x_{k-1}) + M^2_k(x_k-x^*) + \sum_{j=k+1}^n M_j \Delta x_j \end{align}\]

But

\[\begin{align} \begin{split} M^1_k(x^* - x_{k-1}) + M^2_k(x_k-x^*) &\leq M_k (x^* - x_{k-1}) + M_k(x_k-x^*) \\ &=M_k \Delta x_k \end{split} \end{align}\]

Therefore, \(\mathscr{U}(\mathscr{P}^*, f) \leq \mathscr{U}(\mathscr{P}, f)\)

The proof for the lower sum is similar. If $\mathscr{P}^*$ contains $k$ more points than $\mathscr{P}$, we need only repeat the above argument $k$ times to obtain the result. \(\tag*{$\square$}\)

Theorem 6.1.4

Let $f$ be a bounded real-valued function on $[a,b]$. Then

\[\underline{\int_a^b}f \leq \overline{\int_a^b}f.\]

<proof>

Given any two partitions $\mathscr{P}, \mathcal{Q}$ of $[a,b]$. Since $\mathscr{P}\cup\mathcal{Q}$ is both refinement of $\mathscr{P}$ and $\mathcal{Q}$, respectively,

\[\begin{align} \mathscr{L}(\mathscr{P},f) \leq \mathscr{L}(\mathscr{P}\cup\mathcal{Q},f) \leq \mathscr{U}(\mathscr{P}\cup\mathcal{Q},f) \leq \mathscr{U}(\mathcal{Q},f) \end{align}\]

by lemma 6.1.3. Thus $\mathscr{L}(\mathscr{P}, f) \leq \mathscr{U}(\mathcal{Q},f)$ for any partitions $\mathscr{P}, \mathcal{Q}$. Hence

\[\begin{align} \underline{\int_a^b}f = \sup_{\mathscr{P}} \mathscr{L}(\mathscr{P},f) \leq \mathscr{U}(\mathcal{Q},f) \end{align}\]

for any partition $\mathcal{Q}$. Now taking the infimum over $\mathcal{Q}$,

\[\begin{align} \overline{\int_a^b}f= \inf_{\mathcal{Q}}\mathscr{L}(\mathcal{Q},f) \geq \underline{\int_a^b} f \end{align}\]

$\therefore \underline{\int_a^b}f \leq \overline{\int_a^b}f$

Definition 6.1.5

Let $f$ be a bounded real-valued function on the closed and bounded interval $[a,b]$. If

\[\begin{align} \underline{\int_a^b}f = \overline{\int_a^b}f \end{align}\]

then $f$ is said to be Riemann integrable or integrable on $[a,b]$. The common value is denoted by

\[\begin{align} \int_a^b f \quad \text{or } \int_a^b f(x) dx, \end{align}\]

and is called the Riemann Integrable or Integrable of $f$ over $[a,b]$. We denote by $\mathscr{R}[a,b]$ the set of Riemann integrable functions on $[a,b]$. In addition if $f\in\mathscr{R}[a,b]$, we define

\(\int_b^a f :=-\int_a^bf.\) \(\tag*{$\square$}\)

If $f:[a,b]\to\mathbb{R}$ satisfies $m\leq f(t) \leq M$ for all $t\in [a,b]$, then

\[\begin{align} \begin{split} m(b-a)&=\sum_{i=1}^nm(x_i-x_{i-1}) \\ &\leq\sum_{i=1}^nm_i(x_i-x_{i-1})\\ &\leq \underline{\int_a^b} f \\ &\leq \sum_{i=1}^nM_i(x_i-x_{i-1})\\ &\leq \overline{\int_a^b}f \\ &\leq M(b-a). \end{split} \end{align}\]

If, in addition, $f\in \mathscr{R}[a,b]$, then

\[m(b-a) \leq \int_a^b f \leq M(b-a)\]

Example 6.1.6

(a) \(f(x) = \begin{cases} 1 &\quad(x\in\mathbb{Q})\\ 0 &\quad(x\not\in \mathbb{Q}) \end{cases}\)

\[\begin{align} \begin{split} \overline{\int_a^b} f &= \inf_{\mathscr{P}}\mathscr{U}(\mathscr{P},f)=b-a\\ \underline{\int_a^b}f &=\sup_{\mathscr{P}}\mathscr{L}(\mathscr{P},f) = 0 \end{split} \end{align}\]

$\therefore f\not\in \mathscr{R}[a,b]$.

(b) Let \(\mathscr{P} = \{a=x_0, x_1, \ldots, x_n=b\}\) with $x_0<x_1<\cdots <x_n$ be given. Take \(k \in \{1, \ldots, n \}\) satisfying $x_{k-1} < \frac{1}{2} \leq x_k$.

\[\begin{align} \begin{split} M_i &= \begin{cases} 0 &\quad (i=1,\ldots, k-1) \\ 1 &\quad (i=k,\ldots,n) \end{cases} \\ m_i &=\begin{cases} 0 &\quad (i=1,\ldots,k) \\ 1 &\quad (i=k+1,\ldots,n) \end{cases} \end{split} \end{align}\]

Then the upper sum is

\[\begin{align} \begin{split} \mathscr{U}(\mathscr{P},f) &= \sum_{i=1}^n M_i(x_i- x_{i-1}) \\ &= (x_k - x_{k-1}) + \cdots + (x_n -x_{n-1}) \\ &= 1 - x_{k-1} \\ &\geq \frac{1}{2} \end{split} \end{align}\]

For the lower sum,

\[\begin{align} \begin{split} \mathscr{L}(\mathscr{P},f) &= \sum_{i=1}^n m_i (x_i - x_{i-1}) \\ &=1-x_k \\ &\leq \frac{1}{2} \end{split} \end{align}\]

Thus

\[\mathscr{L}(\mathscr{P},f) \leq \underline{\int_0^1} f \leq \frac{1}{2} \leq \overline{\int_a^b}f\leq \mathscr{U}(\mathscr{P},f)\]

Now we want to show that

\[\overline{\int_a^b}f = \underline{\int_a^b} f\]

Note that $\mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) = x_k - x_{k-1}$.

Let $\epsilon >0$ be given. If $\delta x_i = x_i - x_{i-1} < \epsilon$ for all $i$,

\[\begin{align} \begin{split} \overline{\int_0^1}f = \inf_{\mathcal{Q}}\mathscr{U}(\mathcal{Q},f) &\leq \mathscr{U}(\mathscr{P},f) \\ &< \mathscr{L}(\mathscr{P},f) + \epsilon \\ &\leq \sup_{\mathcal{Q}}\mathscr{L}(\mathcal{Q},f) + \epsilon \\ &= \underline{\int_0^1}f + \epsilon \end{split} \end{align}\]

Thus

\[0\leq \overline{\int_a^b}f - \underline{\int_a^b}f < \epsilon\]

for all $\epsilon >0$.

$\therefore$

\[\overline{\int_a^b}f = \underline{\int_a^b}f = \frac{1}{2}\] \[\tag*{$\square$}\]

Theorem 6.1.7

$f$ is a bounded real-valued function on closed and bounded interval $[a,b]$.

\[f\in \mathscr{R}[a,b] \iff \forall \epsilon >0, \exists \mathscr{P}:\text{a partition of }[a,b] \text{ s.t. } \mathscr{U}(\mathscr{P},f) -\mathscr{L}(\mathscr{P},f) < \epsilon\]

Furthermore if $\mathscr{P}$ is a partition with $\mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P}, f) < \epsilon$ for all $\epsilon$, then the equality holds for all the refinement of $\mathscr{P}$.

<proof>

$\Rightarrow$ Suppose that $f\in\mathscr{R}[a,b]$ and let $\epsilon >0$ be given. Since $f$ is integrable, there exist partitions $\mathscr{P_1,P_2}$ of $[a,b]$ such that

\[\begin{align} \int_a^b f -\frac{\epsilon}{2}=\underline{\int_a^b}f -\frac{\epsilon}{2}= \sup_{\mathcal{Q}}\mathscr{L}(\mathcal{Q},f) - \frac{\epsilon}{2} < \mathscr{L}(\mathscr{P}_1,f) \\ \int_a^b f +\frac{\epsilon}{2} = \overline{\int_a^b}f + \frac{\epsilon}{2} = \inf_{\mathcal{Q}}\mathscr{U}(\mathcal{Q},f) + \frac{\epsilon}{2} > \mathscr{U}(\mathscr{P}_2, f) \end{align}\]

That is

\[\int_a^b f - \mathscr{L}(\mathscr{P}_1, f) < \frac{\epsilon}{2}, \quad \mathscr{U}(\mathscr{P}_2, f) -\int_a^b f <\frac{\epsilon}{2}\]

Let $\mathscr{P} = \mathscr{P_1 \cup P_2}$. \(\begin{align} \begin{split} \mathscr{U}(\mathscr{P}, f) &\leq \mathscr{U}(\mathscr{P_2},f ) \\ &< \int_a^b f + \frac{\epsilon}{2} \\ &< \mathscr{L}(\mathscr{P}_1,f) +\epsilon \\& \leq \mathscr{L}(\mathscr{P},f) + \epsilon \end{split} \end{align}\)

Thus we get $\mathscr{U}(\mathscr{P},f ) - \mathscr{L}(\mathscr{P},f ) <\epsilon$.

$\Leftarrow$ Suppose that there is a partition $\mathscr{P}$ of $[a,b]$ such that

\[\mathscr{U}(\mathscr{P},f ) - \mathscr{L}(\mathscr{P},f) < \epsilon.\] \[\begin{align} \begin{split} 0 \leq \overline{\int_a^b}f - \underline{\int_a^b}f &\leq \mathscr{U}(\mathscr{P},f) - \underline{\int_a^b}f \\ &\leq \mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) \\ &<\epsilon \end{split} \end{align}\]

Since $\epsilon >0$ is arbitrary,

\[\overline{\int_a^b}f = \underline{\int_a^b} f\]

$\therefore f\in\mathscr{R}[a,b]$

Now suppose that $\mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) < \epsilon$ and let $\mathscr{P}^*$ be a refinement of $\mathscr{P}$.

Since

\[\mathscr{U}(\mathscr{P}^*,f ) \leq \mathscr{U}(\mathscr{P},f) \text{ and } \mathscr{L}(\mathscr{P}^*, f) \geq \mathscr{L}(\mathscr{P}, f),\]

we get \(\begin{align} \begin{split} \mathscr{U}(\mathscr{P}^*,f) - \mathscr{L}(\mathscr{P}^*,f) &\leq \mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P}^*,f) \\ &\leq \mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) \\ &< \epsilon \end{split} \end{align}\)

\[\tag*{$\square$}\]

Theorem 6.1.8

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

(b) $f$ is monotone on $[a,b] \Rightarrow f\in\mathscr{R}[a,b]$.

<proof>

(a) Let $\epsilon >0$ be given. Choose $\eta>0$ such that

\[(b-a)\eta < \epsilon.\]

Since $f$ is continuous on $[a,b]$, f is uniformly continuous on $[a,b]$. There exists a $\delta>0$ such that

\[|x-t|<\delta \text{ with } x,t \in [a,b] \Rightarrow \left|f(x)-f(t) \right| < \eta.\]

Take a partition $\mathscr{P}$ of $[a,b]$ such that $\Delta x_i < \delta$. Since $f$ is continuous on $[a,b]$, $M_i, m_i \in \text{Range }f$. Thus, $M_i - m_i <\eta$ for all $i.$ Therefore,

\[\begin{align} \begin{split} \mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) &= \sum_{i=1}^n(M_i - m_i)\Delta x_i \\ &< \sum_{i=1}^n \eta\Delta x_i \\ &=(b-a)\eta \\ &< \epsilon \end{split} \end{align}\]

$f\in \mathscr{R}[a,b]$.

(b) Without loss of generality suppose that $f$ is monotone increasing on $[a,b]$. For $n\in\mathbb{N}$, define $h:=(b-a)/n$. For each $i=1,\ldots,n$,

\(x_i := a+ih\) Then \(\mathscr{P}=\{a=x_0, x_1, \ldots, x_n=b\}\) is a partition of $[a,b]$. Let

\[\begin{align} \begin{split} m_i &:=\inf\{f(t): t\in [x_{i-1},x_i]\}\\ M_i &:= \sup\{f(t): t\in [x_{i-1}, x_i]\} \end{split} \end{align}\]

Since $f$ is monotone increasing, $m_i = f(x_{i-1}), M_i = f(x_i)$. Let $\epsilon >0$ be given. We want to show that $\mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) < \epsilon.$

\[\begin{align} \begin{split} \mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) &=\sum_{i=1}^n (M_i - m_i)\Delta x_i \\ &= \sum_{i=1}^n \left(f(x_i) - f(x_{i-1}) \right)h \\ &=h(f(b)-f(a)) \\ &= \frac{b-a}{n}(f(b)-f(a)) \end{split} \end{align}\]

Take $n\in \mathbb{N}$ such that

\[\frac{b-a}{n}(f(b)-f(a)) < \epsilon\]

by Archimedean property.

Then $\mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) <\epsilon$.

$\therefore f\in\mathscr{R}[a.b]$ \(\tag*{$\square$}\)

Reference

• Manfred Stoll, 『Introduction to Real Analysis』, Pearson