Mathematical Analysis
Leture 13
7.1 Partitions, lower and upper integrals
Definition 7.1.1. A partition $P$ of the interval $[a,b]$ is a finite set of numbers $\{ x_0,x_1,x_2,\ldots,x_n \}$ such that \begin{equation*} a = x_0 \lt x_1 \lt x_2 \lt \cdots \lt x_{n-1} \lt x_n = b . \end{equation*} We write \begin{equation*} \Delta x_i := x_i - x_{i-1} . \end{equation*}
7.1 Partitions, lower and upper integrals
Definition 7.1.2. Let $f \colon [a,b] \to \R$ be a bounded function. Let $P$ be a partition of $[a,b]$. Define
$\ds
m_i := \inf \, \bigl\{ f(x) : x_{i-1} \leq x \leq x_i \bigr\} ,
$
$
L(P,f) :=
\ds\sum_{i=1}^n m_i \Delta x_i ,$
We call $L(P,f)$ the lower Darboux sum.
7.1 Partitions, lower and upper integrals
Definition 7.1.2. Let $f \colon [a,b] \to \R$ be a bounded function. Let $P$ be a partition of $[a,b]$. Define
$\ds
M_i := \sup \, \bigl\{ f(x) : x_{i-1} \leq x \leq x_i \bigr\} ,
$
$
U(P,f) :=
\ds\sum_{i=1}^n M_i \Delta x_i .$
We call $U(P,f)$ the upper Darboux sum.
7.1 Partitions, lower and upper integrals
Definition 7.1.2. Let $f \colon [a,b] \to \R$ be a bounded function. Let $P$ be a partition of $[a,b]$. Define
$m_i := \inf \, \bigl\{ f(x) : x_{i-1} \leq x \leq x_i \bigr\} ,$
$L(P,f) := \ds\sum_{i=1}^n m_i \Delta x_i ,$
$M_i := \sup \, \bigl\{ f(x) : x_{i-1} \leq x \leq x_i \bigr\} , $
$U(P,f) := \ds\sum_{i=1}^n M_i \Delta x_i .$
We call $L(P,f)$ the lower Darboux sum and $U(P,f)$ the upper Darboux sum.
Lower and Upper Darboux sums
7.1 Partitions, lower and upper integrals
Theorem 7.1.1. Let $f \colon [a,b] \to \R$ be a bounded function. Let $m, M \in \R$ be such that for all $x \in [a,b]$, we have $m \leq f(x) \leq M$. Then for every partition $P$ of $[a,b]$, \begin{equation} m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a) . \end{equation}
Proof of Theorem 7.1.1
Let $P$ be a partition of $[a,b]$. Note that $m \leq m_i$ for all $i$ and $M_i \leq M$ for all $i,$ and also $m_i \leq M_i$ for all $i$. Finally, $\ds \sum_{i=1}^n \Delta x_i = (b-a)$. Hence
$\,m(b-a) $ $\ds = m \left( \sum_{i=1}^n \Delta x_i \right) $ $\ds = \sum_{i=1}^n m \Delta x_i$ $\ds \leq \sum_{i=1}^n m_i \Delta x_i \quad\quad$
$\ds \leq \ds \sum_{i=1}^n M_i \Delta x_i $ $\ds\leq \sum_{i=1}^n M \Delta x_i $ $\ds = M \left( \sum_{i=1}^n \Delta x_i \right)$ $\ds = M(b-a) .\,\bs$
7.1 Partitions, lower and upper integrals
Definition 7.1.2. As the sets of lower and upper Darboux sums are bounded, we define
$\ds \underline{\int_a^b} f(x)\,dx :=
\sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,b] \bigr\}$
$\ds\overline{\int_a^b} f(x)\,dx :=
\inf \, \bigl\{ U(P,f) : P \text{ a partition of } [a,b] \bigr\} .$
We call $\underline{\int}$ the lower Darboux integral and $\overline{\int}$ the upper Darboux integral.
7.1 Partitions, lower and upper integrals
To avoid worrying about the variable of integration, we often simply write
$\ds \underline{\int_a^b} f \coloneqq \underline{\int_a^b} f(x)\,dx \;$ and $\;\ds \overline{\int_a^b} f \coloneqq \overline{\int_a^b} f(x)\,dx .$
Example 7.1.1.
Consider the Dirichlet function $f \colon [0,1] \to \R,$ where $f(x) \coloneqq 1\,$ if $\,x \in \Q\,$ and $\,f(x) \coloneqq 0\,$ if $\,x \notin \Q.$
Note that for any partition $P$ and every $i$:
$\ds m_i = \inf \bigl\{ f(x) : x \in [x_{i-1},x_i] \bigr\} = 0\;\;$
and $\;\;\ds M_i = \sup \bigl\{ f(x) : x \in [x_{i-1},x_i] \bigr\} = 1 .\qquad\;\,$
Thus $\;\ds L(P,f) $
$= \ds \sum_{i=1}^n 0 \cdot \Delta x_i $
$= 0 \qquad\qquad$
and $\;\;\ds U(P,f) $
$= \ds \sum_{i=1}^n 1 \cdot \Delta x_i $
$= \ds \sum_{i=1}^n \Delta x_i $
$= 1 .$
Example 7.1.1.
Consider the Dirichlet function $f \colon [0,1] \to \R$, where $f(x) \coloneqq 1$ if $x \in \Q$ and $f(x) \coloneqq 0$ if $x \notin \Q$.
Therefore
$\ds \underline{\int_0^1} f = 0\;\;\;$ and $\;\;\;\ds \overline{\int_0^1} f = 1 .$
7.1 Partitions, lower and upper integrals
Definition 7.1.4. Let $P = \{ x_0, x_1, \ldots, x_n \}$ and $\widetilde{P} = \{ \widetilde{x}_0, \widetilde{x}_1, \ldots, \widetilde{x}_{\ell} \}$ be partitions of $[a,b]$.
We say $\widetilde{P}$ is a refinement of $P$ if as sets $P \subset \widetilde{P}$.
7.1 Partitions, lower and upper integrals
Theorem 7.1.2. Let $f \colon [a,b] \to \R$ be a bounded function, and let $P$ be a partition of $[a,b]$. Let $\widetilde{P}$ be a refinement of $P$. Then
\begin{equation*} L(P,f) \leq L(\widetilde{P},f) \;\;\text{and} \;\; U(\widetilde{P},f) \leq U(P,f) . \end{equation*}
7.1 Partitions, lower and upper integrals
Theorem 7.1.3. Let $f \colon [a,b] \to \R$ be a bounded function. Let $m, M \in \R$ be such that for all $x \in [a,b]$, we have $m \leq f(x) \leq M$. Then
\begin{equation} m(b-a) \leq \underline{\int_a^b} f \leq \overline{\int_a^b} f \leq M(b-a) . \end{equation}
Proof of Theorem 7.1.3.
We know that for every partition $P$:
\begin{equation*}
m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a).
\end{equation*}
Then $\;m(b-a) \leq L(P,f)$
$\;\Ra\;$ $\ds m(b-a) \leq \underline{\int_a^b} f,$
and
$\;U(P,f) \leq M(b-a)$
$\;\Ra\;$ $\ds \overline{\int_a^b} f \leq M(b-a)$.
Proof of Theorem 7.1.3.
We know that for every partition $P$: \begin{equation*} m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a). \end{equation*}
$\ds m(b-a) \leq \underline{\int_a^b} f\quad$ $\boxed{?}$ $\quad \ds \overline{\int_a^b} f \leq M(b-a)$.
Consider $P_1, P_2$ partitions of $[a,b]$. Define $\widetilde{P} \coloneqq P_1 \cup P_2$. The set $\widetilde{P}$ is a partition of $[a,b]$, which is a refinement of $P_1$ and a refinement of $P_2$. By Theorem 7.1.2
$\;L(P_1,f) \leq L(\widetilde{P},f)\;$ and $\;U(\widetilde{P},f) \leq U(P_2,f).$
Proof of Theorem 7.1.3.
$\ds m(b-a) \leq \underline{\int_a^b} f\quad$ $\boxed{?}$ $\quad \ds \overline{\int_a^b} f \leq M(b-a)$.
Consider $P_1, P_2$ partitions of $[a,b]$. Define $\widetilde{P} \coloneqq P_1 \cup P_2$. The set $\widetilde{P}$ is a partition of $[a,b]$, which is a refinement of $P_1$ and a refinement of $P_2$. By Theorem 7.1.2
$\;L(P_1,f) \leq L(\widetilde{P},f)\;$ and $\;U(\widetilde{P},f) \leq U(P_2,f).$
$\Ra L(P_1,f) $ $\leq L(\widetilde{P},f) $ $ \leq U(\widetilde{P},f)$ $\leq U(P_2,f) .$
So $\; L(P_1,f) $ $\leq U(P_2,f) .$
Take the supremum and infimum over all partitions!
Proof of Theorem 7.1.3.
Take the supremum and infimum over all partitions!
$\ds \underline{\int_a^b} f $
$\ds =
\sup \, \bigl\{ L(P,f) : P \text{ a partition of } [a,b] \bigr\}\qquad\qquad$
$\quad \ds \leq
\inf \, \bigl\{ U(P,f) : P \text{ a partition of } [a,b] \bigr\}\,$
$\ds =
\overline{\int_a^b} f .$
$\;\bs$
7.2 The Riemann Integral
Definition 7.2.1. Let $f \colon [a,b] \to \R$ be a bounded function such that \begin{equation*} \underline{\int_a^b} f(x)~dx = \overline{\int_a^b} f(x)~dx . \end{equation*} Then $f$ is said to be Riemann integrable.
7.2 The Riemann Integral
The set of Riemann integrable functions on $[a,b]$ is denoted by $\mathcal R[a,b]$. When $f \in \mathcal R[a,b]$, we define \begin{equation*} \int_a^b f(x)~dx := \underline{\int_a^b} f(x)~dx = \overline{\int_a^b} f(x)~dx . \end{equation*} The number $\int_a^b f$ is called the Riemann integral of $f$, or sometimes simply the integral of $f.$
7.2 The Riemann Integral
Theorem 7.2.1 Let $f \colon [a,b] \to \R$ be a Riemann integrable function. Let $m, M \in \R$ be such that $m \leq f(x) \leq M$ for all $x \in [a,b]$. Then \begin{equation*} m(b-a) \leq \int_a^b f \leq M(b-a) . \end{equation*}
Example 7.2.1.
We can use Theorem 7.1.3 to integrate constant functions. If $f(x) \coloneqq c$ for some constant $c$, then we take $m = M = c$. Thus
$\ds m(b-a) \leq \underline{\int_a^b} f \leq \overline{\int_a^b} f \leq M(b-a)$
$\ds c(b-a)= \underline{\int_a^b} f = \overline{\int_a^b} f = c(b-a)$
All the inequalities must be equalities! 😃 Thus $f$ is integrable on $[a,b]$ and \[ \int_a^b f = c\left(b-a\right). \]
7.2 The Riemann Integral
Theorem 7.2.2. Let $f \colon [a,b] \to \R$ be a bounded function. If for every $\epsilon > 0$, there exists a partition $P$ of $[a,b]$ such that \begin{equation*} U(P,f) - L(P,f) \lt \epsilon \end{equation*} then $f$ is Riemann integrable.
Proof. 📝 Exercise.
Example 7.2.2.
Let $f \colon [0,2] \to \R$ be defined by $ f(x) \coloneqq \begin{cases} 1 & \text{if } x \lt 1,\\ 1/2 & \text{if } x = 1,\\ 0 & \text{if } x \gt 1. \end{cases} $
Example 7.2.2.
Claim: $f$ is Riemann integrable and $\ds\int_0^2 f = 1$.
Proof. Consider $0 \lt \epsilon \lt 1$ arbitrary. Let \[ P \coloneqq \{0, 1-\epsilon, 1+\epsilon, 2\} \] be a partition. Using Darboux sums, then
$m_1 = \inf \bigl\{ f(x) : x \in [0,1-\epsilon] \bigr\}$ $\, = 1 ,$
$m_2 = \inf \bigl\{ f(x) : x \in [1-\epsilon,1+\epsilon] \bigr\}$ $\, = 0 ,$
Example 7.2.2.
Proof. Consider $0 \lt \epsilon \lt 1$ arbitrary. Let \[ P \coloneqq \{0, 1-\epsilon, 1+\epsilon, 2\} \] be a partition. Using Darboux sums, then
$m_1 = \inf \bigl\{ f(x) : x \in [0,1-\epsilon] \bigr\}$ $ \,= 1 ,$
$m_2 = \inf \bigl\{ f(x) : x \in [1-\epsilon,1+\epsilon] \bigr\}$ $\, = 0 ,$
$m_3 = \inf \bigl\{ f(x) : x \in [1+\epsilon,2] \bigr\} $ $\, = 0 ,$
Example 7.2.2.
Proof. Consider $0 \lt \epsilon \lt 1$ arbitrary. Let \[ P \coloneqq \{0, 1-\epsilon, 1+\epsilon, 2\} \] be a partition. Using Darboux sums, then
$M_1 = \sup \bigl\{ f(x) : x \in [0,1-\epsilon] \bigr\}$ $\, = 1 ,$
$M_2 = \sup \bigl\{ f(x) : x \in [1-\epsilon,1+\epsilon] \bigr\}$ $ \,= 1 ,$
$M_3 = \sup \bigl\{ f(x) : x \in [1+\epsilon,2] \bigr\} $ $ \,= 0 ,$
Also $\,\Delta x_1 = 1-\epsilon,$ $\,\Delta x_2 = 2\epsilon,\,$ and $\,\Delta x_3 = 1-\epsilon.$
Example 7.2.2.
Proof. Now we compute
$\ds L(P,f) = \sum_{i=1}^3 m_i \Delta x_i \qquad\qquad\qquad\qquad\qquad\qquad\qquad$
$=
1 \cdot (1-\epsilon) $
$
\,+ \,0 \cdot 2\epsilon $
$\,+ \,0 \cdot (1-\epsilon)$
$=1- \epsilon$
$\ds U(P,f) = \sum_{i=1}^3 M_i \Delta x_i \qquad\qquad\qquad\qquad\qquad\qquad\qquad$
$\,=
1 \cdot (1-\epsilon) $
$
\,+ \,1 \cdot 2\epsilon $
$\,+ \,0 \cdot (1-\epsilon)$
$\,=1+ \epsilon$
Example 7.2.2.
Proof. So $ \;L(P,f) = 1 - \epsilon\;$ and $\; U(P,f) = 1+\epsilon. $
Then $\;\ds\overline{\int_0^2} f - \underline{\int_0^2} f $ $\,\leq \, U(P,f) - L(P,f)\qquad \qquad $
$ \qquad \qquad \qquad = (1+\epsilon) \,$ $\,- (1-\epsilon)$ $=2 \epsilon.$
Using Theorem 7.1.3 $\Ra \underline{\int_0^2} f \leq \overline{\int_0^2} f.\,$ Since $\,\epsilon$ was arbitrary,
$\ds \overline{\int_0^2} f = \underline{\int_0^2} f.\;$ So $f$ is Riemann integrable.
Example 7.2.2.
Proof. Finally, we have that
$\ds 1-\epsilon $ $=L(P,f) $ $\ds \leq \int_0^2 f $ $\ds \leq U(P,f) $ $= 1+\epsilon. $
Hence \[ \left|\int_0^2 f - 1 \right| \leq \epsilon. \]
Since $\epsilon$ was arbitrary, we conclude $\ds\int_0^2 f = 1$. $\; \bs$