Properties of the Riemann Integral

Theorem 6.1.9

Let $f:[a,b]\to\mathbb{R}$ be a bounded real valued Riemann integrable function with $Range f \subset [c,d]$. Let $\varphi:[c,d]\to\mathbb{R}$ be a continuous function on $[c,d]$. Then

\[\varphi\circ f \in \mathscr{R}[a,b]\]

<proof> Let $\epsilon >0$ be given. Since $\varphi$ is continuous on $[c,d]$, $\varphi$ is uniformly continuous on $[c,d]$. Let

\[K:=\sup \{ |\varphi(t)|: t\in [c,d]\}\]

and let

\[\epsilon^\prime := \frac{\epsilon}{b-a+2K}.\]

Since $\varphi$ is uniformly continuous on $[c,d]$, there exists $0<\delta<\epsilon^\prime$ such that

\[|s-t|<\delta \text{ with } s,t\in[c,d] \Rightarrow |\varphi(s)-\varphi(t)|<\epsilon^\prime.\]

Since $f\in\mathscr{R}[a,b]$, there is a partition

\[\mathscr{P}=\{x_0, \ldots, x_n\} \text{ of } [a,b] \text{ such that } \mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) < \delta^2.\]

Now we want to show that $\mathscr{U}(\mathscr{P}, \varphi\circ f) - \mathscr{L}(\mathscr{P},\varphi\circ f) < \epsilon$. We write

\[\begin{align} \begin{split} m_k &= \inf \{ f(t): t\in [x_{k-1}, x_k]\}\\ M_k &=\sup \{f(t):t\in [x_{k-1}, x_k]\} \\ m^*_k &= \inf \{ (\varphi\circ f)(t): t\in[x_{k-1},x_k]\} \\ M^*_k &= \sup \{ (\varphi\circ f)(t): t\in[x_{k-1},x_k]\} \end{split} \end{align}\]

Let

\[\begin{align} \begin{split} A&:=\{k: M_k - m_k <\delta\}\\ B &:=\{k: M_k - m_k \geq \delta \} \end{split} \end{align}\]

If $k\in A$, then

\[\left|\varphi(f(s))-\varphi(f(t)) \right| < \epsilon^\prime\]

for all $s,t\in [x_{k-1}, x_k]$. For $k\in A$,

\[\begin{align} \begin{split} M^*_k - m^*_k &=\sup\{\varphi(f(t)): t \in [x_{k-1}, x_k]\} - \inf\{\varphi(f(t)): t \in [x_{k-1}, x_k]\} \\ &=\sup \{\varphi(f(t)): t \in [x_{k-1}, x_k]\} + \sup \{-\varphi(f(t)): t \in [x_{k-1}, x_k]\} \\ &=\sup\{\varphi(f(t))-\varphi(f(s)): t,s \in [x_{k-1}, x_k]\} \\ &\leq \sup\{\left|\varphi(f(t))-\varphi(f(s))\right|: t,s\in [x_{k-1},x_k]\}\\ &\leq \epsilon^\prime \end{split} \end{align}\]

If $k\in B$,

\[\begin{align} \begin{split} M^*_k &=\sup\{\varphi(f(t)): t\in [x_{k-1},x_k]\} \\ &\leq \sup \{|\varphi(f(t))|: t\in [x_{k-1},x_k]\} \\ &\leq \sup \{|\varphi(f(t))|: t\in [c,d]\} \\ &=K \end{split} \end{align}\]

Similarly,

\[\begin{align} \begin{split} -m^*_k &= \sup\{-\varphi(f(t)): t\in [x_{k-1},x_k]\}\\ &\leq \sup\{ |\varphi(f(t)): t\in [x_{k-1},x_k]\}\\ &\leq \sup \{|\varphi(f(t))|: t\in [c,d]\} \\ &=K \end{split} \end{align}\]

So, \(M^*_k -m^*_k \leq 2K\).

Now, \(\begin{align} \begin{split} \mathscr{U}(\mathscr{P}, \varphi\circ f) - \mathscr{L}(\mathscr{P},\varphi\circ f) &= \sum_{i=1}^n (M^*_i- m^*_i)\Delta x_i \\ &=\sum_{k\in A}(M^*_k- m^*_k)\Delta x_k + \sum_{k\in B}(M^*_k- m^*_k)\Delta x_k \\ &\leq \epsilon^\prime \sum_{k\in A}\Delta x_k + 2K\sum_{k\in B} \Delta x_k \\ &< \epsilon^\prime (b-a) + 2K \sum_{k\in B} \Delta x_k \end{split} \end{align}\)

Then we need to derive the upper bound of sum of $\Delta x_k$ with the condition $f \in\mathscr{R}[a,b]$ and $M_k - m_k \geq \delta$ for $k\in B$.

Since $M_k - m_k \geq \delta$ for $k\in B$, $\frac{M_k-m_k}{\delta} \geq 1$.

\[\begin{align} \begin{split} \sum_{k\in B} \Delta x_k &\leq \sum_{k\in B} \frac{M_k-m_k}{\delta}\Delta x_k \\ &\leq \frac{1}{\delta}\sum_{i=1}^n(M_i-m_i)\Delta x_i\\ &=\frac{1}{\delta}\left(\mathscr{U}(\mathscr{P},f) - \mathscr{L}(\mathscr{P},f) \right) \\ &< \delta \\ &< \epsilon^\prime \end{split} \end{align}\]

Combining the equation $(6)$ and $(7)$, we get

\[\begin{align} \begin{split} \mathscr{U}(\mathscr{P}, \varphi\circ f) - \mathscr{L}(\mathscr{P}, \varphi\circ f) &< \epsilon^\prime (b-a) + 2K \epsilon^\prime\\ &=\epsilon^\prime (b-a+2K)\\ &=\epsilon \end{split} \end{align}\]

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

Theorem 6.2.1

Let $f,g$ be bounded real-valued functions on $[a,b]$ and $f,g,\in\mathscr{R}[a,b]$.

$(a) f+g\in\mathscr{R}[a,b]$ with $\int_a^b(f+g) = \int_a^b f+ \int_a^b g$.

$(b) cf \in \mathscr{R}[a,b]$ with $\int_a^b cf = c\int_a^b f$.

$(c) fg \in \mathscr{R}[a,b]$.

<proof>

$(a)$ Let

\[\mathscr{P} = \{x_0, \ldots, x_n\}\]

be a partition of $[a,b]$. Then

\[\mathscr{U}(\mathscr{P}, f+g) \leq \mathscr{U}(\mathscr{P}, f) + \mathscr{U}(\mathscr{P}, g)\]

$\because$

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

For $t\in [x_{i-1}, x_i]$, $f(t)+g(t) \leq M_i(f) + M_i(g)$. So,

\[\sup \{f(t)+g(t): t\in [x_{i-1},x_i]\} \leq M_i(f) + M_i(g)\]

Thus,

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

Let $\epsilon >0$ be given. Since $f,g\in \mathscr{R}[a,b]$, there are partitions of $\mathscr{P}_f, \mathscr{P}_g$ of $[a,b]$, respectively such that

\[\begin{align} \begin{split} \mathscr{U}(\mathscr{P}_f, f) &<\inf_{\mathcal{Q}}\mathscr{U}(\mathcal{Q},f) +\frac{\epsilon}{2}= \overline{\int_a^b}f + \frac{\epsilon}{2} = \int_a^b f + \frac{\epsilon}{2} \\ \mathscr{U}(\mathscr{P}_g, g) &<\inf_{\mathcal{Q}}\mathscr{U}(\mathcal{Q},g) +\frac{\epsilon}{2}= \overline{\int_a^b}g + \frac{\epsilon}{2} = \int_a^b g + \frac{\epsilon}{2} \end{split} \end{align}\]

Let $\mathcal{Q} = \mathscr{P}_f \cup \mathscr{P}_g$. Then $\mathcal{Q}$ is a refinement of $\mathscr{P}_f$ and $\mathscr{P}_g$, respectively.

\[\begin{align} \begin{split} \overline{\int_a^b}f+g &\leq \mathscr{U}(\mathcal{Q}, f+g) \\ &\leq \mathscr{U}(\mathcal{Q},f) + \mathscr{U}(\mathcal{Q},g)\\ &=\mathscr{U}(\mathscr{P}_f, f) + \mathscr{U}(\mathscr{P}_g, g) \\ &<\int_a^bf + \int_a^bg+\epsilon \end{split} \end{align}\]

So, we get

\[\overline{\int_a^b}(f+g) \leq \int_a^b f + \int_a^b g + \epsilon.\]

Since $\epsilon$ is arbitrary, we have

\[\overline{\int_a^b}(f+g) \leq \int_a^b f + \int_a^b g\]

Similarly, $\mathscr{L}(\mathscr{P}, f+g) \geq \mathscr{L}(\mathscr{P},f) + \mathscr{L}(\mathscr{P}, g)$.

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

Since $f\in\mathscr{R}[a,b]$, there exist partition $\mathscr{P}_f, \mathscr{P}_g$ of $[a,b]$, respectively such that

\[\begin{align} \mathscr{L}(\mathscr{P}_f, f) + \frac{\epsilon}{2} &> \sup_{\mathcal{Q}}\mathscr{L}(\mathcal{Q}, f) = \underline{\int_a^b}f = \int_a^b f\\ \mathscr{L}(\mathscr{P}_g, g) + \frac{\epsilon}{2} &> \sup_{\mathcal{Q}}\mathscr{L}(\mathcal{Q}, g) = \underline{\int_a^b}g = \int_a^b g \end{align}\]

Let $\mathscr{S}= \mathscr{P}_f \cup \mathscr{P}_g$. Then $\mathscr{S}$ is a refinement of $\mathscr{P}_f$ and $\mathscr{P}_g$, respectively.

\[\begin{align} \begin{split} \underline{\int_a^b}f+g &\geq \mathscr{L}(\mathscr{S}, f+g) + \mathscr{L}(\mathscr{S}, g) \\ &\geq \mathscr{L}(\mathscr{P}_f, f) + \mathscr{L}(\mathscr{P}_g, g)\\ &> \int_a^b f + \int_a^b g - \epsilon \end{split} \end{align}\]

Since $\epsilon >0$ is arbitrary, we get

\[\underline{\int_a^b}f+g \geq \int_a^b f + \int_a^b g\]

Finally,

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

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

\[\int_a^b f+g = \int_a^b f + \int_a^b g\] \[\tag*{$\square$}\]

Lemma 1

Let $A$ be a bounded subset of $\mathbb{R}$. For a constant $c\in\mathbb{R}$, define

\[cA :=\{ca: a\in A\}.\]

$(a)$ If $c>0, \sup(cA) = c\sup A$, and $\inf(cA) = c\inf A$

$(b)$ If $c<0, \sup(cA) = c\inf A$, and $\inf(cA) = c\sup A$.

<proof>

Suppose that $c>0$. Let $M:=\sup A$ and let $\epsilon >0$ be given. Then there exists a $a_\epsilon \in A$ such that

\[M-\frac{\epsilon}{c} < a_\epsilon \leq M\]

Since $c>0$,

\[cM-\epsilon < ca_\epsilon \leq cM\]

Moreover $cM$ is an upper bound of $cA$ since $ca \leq cM$ for all $a\in A$.

$\therefore cM=\sup(cA)$ by theorem 1.4.4.

Similarly, let $L :=\inf A$. Then there exists a $a_\epsilon \in A$ such that

\[\begin{align} \begin{split} L&\leq a_\epsilon < L+\frac{\epsilon}{c} \\ cL &\leq ca_\epsilon < L + \epsilon \end{split} \end{align}\]

Moreover, $cL$ is a lower bound of $cA$.

$\therefore cL = \inf(cA)$.

$(b)$ Suppose that $c<0$. Let $\epsilon >0$ be given. SInce $-\epsilon/c >0$, there is a $a_\epsilon \in A$ such that

\[\begin{align} \begin{split} M+\frac{\epsilon}{c} &< a_\epsilon \leq M\\ cM &\leq ca_\epsilon < cM +\epsilon \end{split} \end{align}\]

Since $c<0$ and $a \leq M$, $cM \leq a$ for all $a\in A$.

$\therefore cM=\inf(cA)$.

Similarly, there is a $a^\prime_\epsilon \in A$ such that

\[\begin{align} \begin{split} L&\leq a^\prime_\epsilon < L -\frac{\epsilon}{c} \\ cL-\epsilon&<a^\prime_\epsilon \leq cL \end{split} \end{align}\]

$\therefore \sup(cA) = cL = c\inf A$.

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

$(b)$

<proof>

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

Suppose that $c>0$. Let

\[\mathscr{P}=\{x_0,\ldots, x_n\}\]

be a partition of $[a,b]$.

By lemma1,

\[\begin{align} M_i(cf) &=cM_i(f) \\ m_i(cf) &=cm_i(f) \end{align}\]

Then

\[\begin{align} \begin{split} \mathscr{U}(\mathscr{P},cf) &=\sum_{i=1}^n M_i(cf)\Delta x_i\\ &=c\sum_{i=1}^n M_i(f) \Delta x_i\\ &=c\mathscr{U}(\mathscr{P},f) \\ &= c \overline{\int_a^b}f \\ &=c\int_a^b f \: (\because f \in\mathscr{R}[a,b]) \end{split} \end{align}\] \[\begin{align} \begin{split} \overline{\int_a^b}cf &= \inf_{\mathcal{Q}}\mathscr{U}(\mathcal{Q},cf) \\ &=\inf_{\mathcal{Q}}c \mathscr{U}(\mathcal{Q},f) \\ &=c\inf_{\mathcal{Q}} \mathscr{U}(\mathcal{Q},f) \:(\because \text{ by lemma}1)\\ &=c\overline{\int_a^b}f \\ &=c\int_a^b f \end{split} \end{align}\]

Similarly, we get

\[\begin{align} m_i(cf) &= cm_i(f) \\ \mathscr{L}(\mathscr{P},cf) &= c \mathscr{L}(\mathscr{P},f). \end{align}\]

So,

\[\begin{align} \begin{split} \underline{\int_a^b}cf &= \sup_{\mathcal{Q}}\mathscr{L}(\mathcal{Q}, cf)\\ &=c\sup_{\mathcal{Q}}\mathscr{L}(\mathcal{Q},f) \:(\because \text{ by lemma}1) \\ &=c\underline{\int_a^b}f \\ &=c{\int_a^b}f \end{split} \end{align}\]

$\therefore$

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

For $c=0$,

\[\mathscr{U}(\mathscr{P},cf) = \mathscr{L}(\mathscr{P},cf)=0\]

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

\[\int_a^b cf = 0 =c\int_a^b f\]

Now, suppose that $c<0$.

\(\begin{align} \begin{split} M_i(cf) &= \sup\{ cf(t): t\in [x_{i-1},x_i]\} \\ &=c\inf\{f(t):t\in [x_{i-1},x_i]\} \:(\because \text{ by lemma}1)\\ &=cm_i(f) \end{split} \end{align}\) Now, we get

\[\begin{align} \begin{split} \mathscr{U}(\mathscr{P}, cf) &= \sum_{i=1}^n M_i(cf)\Delta x_i\\ &=\sum_{i=1}^n cm_i(f)\Delta x_i\\ &= c\mathscr{L}(\mathscr{P},f) \end{split} \end{align}\] \[\begin{align} \begin{split} \overline{\int_a^b}cf &=\inf_{\mathcal{Q}}\mathscr{U}(\mathcal{Q},f) \\ &=\inf_{\mathcal{Q}}c\mathscr{L}(\mathcal{Q},f) \\ &=c\sup_{\mathcal{Q}}\mathscr{L}(\mathcal{Q},f) \\ &= \underline{\int_a^b}f \\ &= c\int_a^b f \end{split} \end{align}\]

Similarly, we can derive that

\[\begin{align} m_i(cf) &= cM_i(f) \\ \mathscr{L}(\mathscr{P},cf) &= c\mathscr{U}(\mathscr{P},f) \end{align}\]

So,

\[\begin{align} \begin{split} \underline{\int_a^b}cf &=\sup_{\mathcal{Q}}\mathscr{L}(\mathcal{Q}, cf) \\ &= \sup_{\mathcal{Q}} c\mathscr{U}(\mathcal{Q},f) \\ &=c\inf_{\mathcal{Q}}\mathscr{U}(\mathcal{Q},f) \\ &=c \overline{\int_a^b}f \\ &= c\int_a^b f \end{split} \end{align}\]

Thus, we get

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

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

\[\int_a^b cf = c\int_a^bf\]

$(c)$

<proof>

\[fg = \frac{1}{4}\left[(f+g)^2 - (f-g)^2 \right]\]

By $(a)$, $f+g\in\mathscr{R}[a,b]$ and $-1\cdot g\in\mathscr{R}[a,b]$ by $(c)$. Consequently, $f-g\in\mathscr{R}[a,b]$. Since $y=x^2$ is continuous function, $(f+g)^2, (f-g)^2\in\mathscr{R}[a,b]$ by theorem 6.1.9.

Similarly, $(f+g)^2 - (f-g)^2\in\mathscr{R}[a,b]$. Finally $\frac{1}{4}\left[(f+g)^2 - (f-g)^2 \right]\in\mathscr{R}[a,b]$.

\(\tag*{$\square$}\)

Reference

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