3.26 The properties of Integral

Definition of the Riemann Integral

In the previous lecture, we have defined the Riemann integral for any bounded function \(f: [a, b] \to \mathbb{R}\) on a closed interval \([a, b]\); here we were implicitly assuming that \(a < b\).

In case the function \(f\) is integrable, that is, \(L(f)=U(f)\)
We write \(\displaystyle\int_{a}^{b}f(x)dx=L(f)=U(f).\)

What if \(a > b\) or \(a = b\)?

To make sense with the definition above, and with the idea of the relation of the integral with computing areas, we define:

Definition

Let \(f: [a, b] \to \mathbb{R}\) be a bounded function which is integrable on \([a, b]\).

We set \(\displaystyle\int_{a}^{a} f(x)dx = 0\) and \(\displaystyle\int_{b}^{a} f(x)dx = -\int_{a}^{b} f(x)dx.\)

Properties of the Riemann Integral

The properties of the Riemann integral concern two features:

• The intervals related to the domain of the function.
• Some algebraic properties of the functions we are integrating.

Theorem: Splitting the Integral

Let \(f: [a, b] \to \mathbb{R}\) be a bounded function defined on a closed interval \([a, b]\) and let \(c \in [a, b]\).

Then \(f\) is integrable on \([a, b]\) if and only if it is integrable on \([a, c]\) and \([c, b]\).

In such a case, we have \(\displaystyle\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx.\)

Proof

\(\Rightarrow\)) Suppose \(f\) is integrable in \([a,b]\), then \(U(f)=L(f)=\displaystyle \int^{b}_{a}f(x)dx\)

We know \(\forall\varepsilon>0,\exists P_{\varepsilon}\in\mathcal{P}:0<U(f,P_{\varepsilon} )-L(f,P_{\varepsilon})<\varepsilon\)

Here \(P_{\varepsilon}=(x_{0}=a,x_{1},...,x_{n}=b)\), then \(c\) is in \(P_\varepsilon\) or not

• If \(c\in P_\varepsilon\)

Then \(c=x_k\) for some \(k\), then \(L(f,P_{\varepsilon})=\sum\limits_{i=1}^{n}m_{i}(x_{i}-x_{i-1})=\sum\limits_{i=1} ^{k}m_{i}(x_{i}-x_{i-1})+\sum\limits_{i=k+1}^{n}m_{i}(x_{i}-x_{i-1})\)

And let \(P_1=(x_0=a,...,x_k=c)\) and \(P_2=(x_{k+1},...,x_{n}=b)\)

Then \(L(f,P_{\varepsilon})=L\left(f,P_{1}\right)+L\left(f,P_{2}\right)\), similarly \(U(f,P_{\varepsilon})=U\left(f,P_{1}\right)+U\left(f,P_{2}\right)\)

Thus \(0\leq U(f,P_{\varepsilon})-L(f,P_{\varepsilon})=U\left(f,P_{1}\right)+U\left(f,P _{2}\right)-\left(L\left(f,P_{1}\right)+L\left(f,P_{2}\right)\right)\)

\(=\underbrace{U\left(f,P_{1}\right)-L\left(f,P_{1}\right)}_{\geq 0}+\underbrace{U\left(f,P_{2}\right)-L\left(f,P_{2} \right)}_{\geq 0}<\varepsilon\)

Thus \(0\leq U(f,P_{1})-L(f,P_{1})<\varepsilon\) and \(0\leq U(f,P_{2})-L(f,P_{2})<\varepsilon\)

Thus \(f\) is integrable in \([a,c]\) and \([c,b]\) * If \(c\notin P_\varepsilon\)​

We consider another partition by adding \(c\), which is \(Q_\varepsilon =(x_0=a,...,x_k,c,x_{k+1},...,x_n=b)\) contains \(P_\varepsilon\)​

Thus \(Q_\varepsilon\) is a refinement, then \(U(f,Q_\varepsilon)\leq U(f,P_\varepsilon)\) and \(L(f,Q_{\varepsilon})\geq L(f,P_{\varepsilon})\)

But \(0\leq U(f,Q_{\varepsilon})-L(f,Q_{\varepsilon})\leq U(f,P_{\varepsilon})-L(f,P_{\varepsilon} )<\varepsilon\)​

By what we did before, this implies that there exists partitions \(Q_1\) of \([a,c]\) and \(Q_2\) of \([c,b]\) such that \(0\leq U(f,Q_{1})-L(f,Q_{1})<\varepsilon\) and \(0\leq U(f,Q_{2})-L(f,Q_{2})<\varepsilon\)

Thus \(f\) is integrable in \([a,c]\) and \([c,b]\)

\(\Leftarrow\)) Suppose \(f\) is integrable in \([a,c]\) and \([c,b]\). NTP \(f\) is integrable in \([a,b]\)

We know \(\exists P_1\) of \([a,c]\) and \(P_2\) of \([c,b]\) such that \(0\leq U(f,P_{1})-L(f,P_{1})<\varepsilon\) and \(0\leq U(f,P_{2})-L(f,P_{2})<\varepsilon\)

Consider \(P_1\cup P_2\) of \([a,b]\), then \(U(f,P_1\cup P_2)\leq U(f,P_1),U(f,P_2)\) and \(L(f,P_{1}\cup P_{2})\geq L(f,P_{1}),L(f,P_{2})\)

Then we have \(0\leq U(f,P_{1}\cup P_{2})-L(f,P_{1}\cup P_{2})<U(f,P_{1})-L(f,P_{1})<\varepsilon\)

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Then we prove that we can sum the integrals.

Since \(\displaystyle \int^{b}_{a}f(x)dx=U(f)\leq U(f,P),\forall P\in \mathcal{P}\)

Let's take the partition \(P_{\varepsilon}=P_{1}\cup P_{2}\), then \({\displaystyle\int_{a}^{b}f(x)dx\leq U(f,P_{\varepsilon})=U\left(f,P_1\right)+U\left(f,P_2\right)}\)

And we know \(L(f,P_1)\leq L(f)=\displaystyle \int^c_af(x)dx\leq U(f)\leq U(f,P_1)\) and \(U(f,P_{1})\leq\frac{\varepsilon}{2}+L(f,P_{1})\)

\(L(f,P_{2})\leq L(f)={\displaystyle\int_{c}^{b}f(x)dx\leq U(f)\leq U(f,P_2)}\) and \(U(f,P_{2})\leq\frac{\varepsilon}{2}+L(f,P_{2})\)

Then \({\displaystyle\int_{a}^{b}f(x)dx\leq\varepsilon+L\left(f,P_1\right)+L\left(f,P_2\right)\leq\varepsilon+\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx}\)

Then \({\displaystyle\int_{a}^{b}f(x)\,dx\leq\int_{a}^{c}f(x)\,dx+\int_{c}^{b}f(x)\,dx}\)

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Now

\({{\displaystyle\int_{a}^{c}f(x)\,dx+\int_{c}^{b}f(x)\,dx\leq}}\frac{\varepsilon}{2} +L(f,P_{1})+\frac{\varepsilon}{2}+L(f,P_{2})\leq\varepsilon+(L(f,P_{1})+L(f,P_{2} ))\)

\({\displaystyle\leq\varepsilon+L(f,P_{\varepsilon})\leq\varepsilon+\int_{a}^{b}f(x)\,dx,\forall\varepsilon>0}\)

Thus \(\displaystyle\int_{a}^{c}f(x)\,dx+\int_{c}^{b}f(x)\,dx\leq\int_{a}^{b}f(x)\,dx\)

Thus \(\displaystyle\int_{a}^{b} f(x) \,dx = \int_{a}^{c} f(x) \,dx + \int_{c}^{b} f(x) \,dx\)

Algebraic Properties of the Integral

Let \(f\) and \(g\) be two integrable functions on the closed interval \([a, b]\). Then:

1. Sum Property: The function \(f + g\) is integrable on \([a, b]\), and \(\displaystyle\int_{a}^{b} (f(x) + g(x))dx = \int_{a}^{b} f(x)dx + \int_{a}^{b} g(x)dx.\)

   Proof

   Let \(P=(x_0,...,x_n)\) be partition on \([a,b]\), then \(L(f,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right)m_{i}\) and \(U(f,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right)M_{i}\)

   Also, \(m_{i}=\inf\{f(x)\mid x\in[x_{i-1},x_{i}]\}\), \(m_{i}^{\prime}=\inf\{g(x)\mid x\in[x_{i-1},x_{i}]\}\)

   Then \(m_{i}+m_{i}^{\prime}=\inf\{f(x)\mid x\in[x_{i-1},x_{i}]\}+\inf\{g(x)\mid x\in[x_{i-1} ,x_{i}]\}\leq\inf\{\left(f+g)(x\right)\mid x\in[x_{i-1},x_{i}]\}=m_{i}^{*}\)

   Then \(L(f+g,P)=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)m_{i}^{*}\geq\sum_{i=1}^{n}\left (x_{i}-x_{i-1}\right)\left(m_{i}+m_{i}^{\prime}\right)=L(f,P)+L(g,P)\)

   Similarly, \(U(f+g,P)\leq U(f,P)+U(g,P)\)

   Then \(U(f+g,P_{n})-L(f+g,P_{n})\leq U(f,P_{n})+U(g,P_{n})-L(f,P_{n})-L\left(g,P_{n}\right )=U\left(f,P_{n}\right)-L\left(f,P_{n}\right)+U\left(g,P_{n}\right)-L\left(g,P_{n} \right)\)

   Since \(f,g\) is integrable, then \(U(f,P_{n})-L(f,P_{n})<\frac{\varepsilon}{2},U(g,P_{n})-L(g,P_{n})<\frac{\varepsilon}{2}\)

   Thus \(U(f+g,P_{n})-L(f+g,P_{n})<\varepsilon\). Thus \(f+g\) is integrable.

   And \({\displaystyle\int_{a}^{b}(f+g)(x)dx=\lim_{n\to\infty}U(f+g,P_{n})}\) \({\displaystyle\leq\lim_{n\to\infty}U(f,P_{n})+\lim_{n\to\infty}U(g,P_{n})=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx}\)

   \({{\displaystyle\int_{a}^{b}(f+g)(x)dx=\lim_{n\to\infty}L(f+g,P_{n})}}\) \({{\displaystyle\geq\lim_{n\to\infty}L(f,P_{n})+\lim_{n\to\infty}L(g,P_{n})=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx}}\)

   Thus \(\displaystyle\int_{a}^{b}(f+g)(x)dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx\)

2. Scalar Multiplication Property: For any \(c \in \mathbb{R}\), the function \(cf\) is integrable on \([a, b]\), and \(\displaystyle\int_{a}^{b} cf(x)dx = c \int_{a}^{b} f(x)dx.\)

   Proof

   Let \(P = (x_0, x_1, \dots, x_n)\) be a partition of \([a, b]\).

   So, for \((kf)(x) = k \cdot f(x)\), \(\forall x \in [a, b]\):

   \(m_{i}' = \inf \{ (kf)(x) \mid x \in [x_{i-1}, x_{i}] \}\) \(= \inf \{ k f(x) \mid x \in [x_{i-1}, x_i] \}\) \(= k \inf \{ f(x) \mid x \in [x_{i-1}, x_i] \}\) \(=km_{i},\forall i\)

   Hence, \(L(kf, P) = \sum_{i=1}^{n} m_i' (x_i - x_{i-1})\) \(= \sum_{i=1}^{n} k m_i (x_i - x_{i-1})\) \(= k \sum_{i=1}^{n} m_i (x_i - x_{i-1})\) \(=kL(f,P),\forall P\in P.\)

   Analogously, \(U(kf, P) = k U(f, P)\)​

   Since \(f\) is integrable, \(\exists(P_{n})_{n\in\mathbb{N}}\) sequence of partitions such that \(\lim_{n \to \infty}U(f, P_{n}) - L(f, P_{n}) = 0\) and \({\displaystyle\int_{a}^{b}f(x)dx=\lim_{n\to\infty}U(f,P_{n})}=\lim_{n\to\infty}L (f,P_{n})\)​

   Now, \(\lim_{n \to \infty} U(kf, P_n) - L(kf, P_n)\) \(= \lim_{n \to \infty} k U(f, P_n) - k L(f, P_n)\) \(= \lim_{n \to \infty}k (U(f, P_{n}) - L(f, P_{n}))\)

   \(=k\lim_{n\to\infty}(U(f,P_{n})-L(f,P_{n}))=0\). Thus \(kf\) is integrable.

   And \(\displaystyle\int_{a}^{b}(kf)(x) dx = \lim_{n \to \infty}U(kf, P_{n})\) \(\displaystyle = k \lim_{n \to \infty}U(f, P_{n}) = k \int_{a}^{b}f(x)dx\)

   Remark: The properties (1) and (2) are called the linearity of the integral.

3. Bounded Function Property:
   If \(m \leq f(x) \leq M\) for all \(x \in [a, b]\), then \(\displaystyle m(b - a) \leq \int_{a}^{b} f(x)dx \leq M(b - a).\)

   ​

   Proof

   Let \(P = (x_0, x_1, \dots, x_n)\) be a partition.

   Then \(m_{i}=\left\lbrace\inf f(x),x\in[x_{i-1},x_{i}]\right\rbrace\geq m\text{ for all }i\) and \(M_{i}=\left\lbrace\sup f(x),x\in[x_{i-1},x_{i}]\right\rbrace\leq M\text{ for all }i\)

   \(L(f,P)=\sum_{i=1}^{n}m_{i}(x_{i}-x_{i-1})\geq\sum_{i=1}^{n}m(x_{i}-x_{i-1})= m \sum_{i=1}^{n}(x_{i} - x_{i-1}) = m (b - a)\)

   Then \(m(b - a) \leq L(f,P) \leq L(f)\)

   Similarly, \(U(f) \leq U(f,P) \leq M(b - a)\)

   Since \(f\) is integrable, \(L(f) = U(f)\)​\(\displaystyle\Rightarrow m(b - a) \leq \int_{a}^{b} f(x) dx \leq M(b - a)\)​

4. Order Property: If \(f(x) \leq g(x)\) for all \(x \in [a, b]\), then \(\displaystyle\int_{a}^{b} f(x)dx \leq \int_{a}^{b} g(x)dx.\)

   \(\text{If }f(x)\leq g(x),\forall x\in[a,b]\) \(\Rightarrow0\leq g(x)-f(x),\forall x\in[a,b]\)

   By (3), \(\displaystyle0 \cdot (b - a) \leq \int_{a}^{b} (g(x) - f(x)) \,dx\) \(\displaystyle \Rightarrow 0 \leq \int_{a}^{b} \left( g(x) + (-f(x)) \right) \,dx\)

   By (1), \(\displaystyle 0 \leq \int_{a}^{b} g(x) \,dx + \int_{a}^{b} (-f(x)) \,dx\) \(\displaystyle = \int_{a}^{b} g(x) \,dx + \int_{a}^{b} (-1) f(x) \,dx\) \(\displaystyle \Rightarrow 0 \leq \int_{a}^{b} g(x) \,dx - \int_{a}^{b} f(x) \,dx\) by (2)

   Then \(\displaystyle\int_{a}^{b}f(x)\,dx\leq\int_{a}^{b}g(x)\,dx\)

5. Absolute Value Property: The function \(|f|\) is integrable on \([a, b]\), and \(\displaystyle\left| \int_{a}^{b} f(x)dx \right| \leq \int_{a}^{b} |f(x)|dx.\)

   Let \(P=(x_0,...,x_n)\) be partition on \([a,b]\), then \(L(\left|f\right|,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right)m_{i}\) and \(U(|f|,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right)M_{i}\)

   Also, \(m_{i}=\inf\{\left|f(x)\right|\mid x\in[x_{i-1},x_{i}]\}\) and \(M_{i}=\sup\{\left|f(x)\right|\mid x\in[x_{i-1},x_{i}]\}\)

   \(m'_{i}=\inf\{ f(x) \mid x\in[x_{i-1},x_{i}]\}\) and \(M_{i}=\sup\{ f(x) \mid x\in[x_{i-1},x_{i}]\}\)

   Then since \(|f(x)-f(y)|\geq ||f(x)|-|f(y)||\), then \(M_i-m_i\leq M_i'-m_i'\)

   Then \(U(\left|f\right|,P)-L(\left|f\right|,P)=\sum_{i=0}^{n}\left(x_{i}-x_{i-1}\right) \left(M_{i}-m_{i})\leq U(f,P\right)-L(f,P)\)

   Since \(f\) is integrable, then \(U(f,P)-L(f,P)<\varepsilon\)

   Thus \(U(\left|f\right|,P_{n})-L(\left|f\right|,P_{n})<\varepsilon\). Thus \(|f|\) is integrable.

   ---

   We know \(f(x)\leq |f(x)|\), then by 4, we have \({\displaystyle\int_{a}^{b}f(x)dx\leq\int_{a}^{b}|f(x)|dx.}\)​

   Since \(-|f(x)|\leq f(x)\), then by 4, we have \(\displaystyle\int_{a}^{b}-|f(x)|dx\leq \int_{a}^{b}f(x)dx\Rightarrow -\int_{a}^{b}|f(x)|dx\leq \int_{a}^{b}f(x)dx\)

   Thus we have \(\displaystyle\left| \int_{a}^{b} f(x)dx \right| \leq \int_{a}^{b} |f(x)|dx.\)

Remark

The properties 1) and 2) are called the linearity of the integral.