7.3: Integrability Conditions
- Page ID
- 22680
If \(a<b\) and \(f:[a, b] \rightarrow \mathbb{R}\) is monotonic, then \(f\) is integrable on \([a, b]\).
- Proof
-
Suppose \(f\) is nondecreasing. Given \(\epsilon>0,\) let \(n \in Z^{+}\) be large enough that
\[\frac{(f(b)-f(a))(b-a)}{n}<\epsilon .\]
For \(i=0,1, \ldots, n,\) let
\[x_{i}=a+\frac{(b-a) i}{n}.\]
Let \(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\} .\) Then
\[\begin{aligned} U(f, P)-L(f, P) &=\sum_{i=1}^{n} f\left(x_{i}\right)\left(x_{i}-x_{i-1}\right)-\sum_{i=1}^{n} f\left(x_{i-1}\right)\left(x_{i}-x_{i-1}\right) \\ &=\sum_{i=1}^{n}\left(f\left(x_{i}\right)-f\left(x_{i-1}\right)\right) \frac{b-a}{n} \\ &=\frac{b-a}{n}\left(\left(f\left(x_{1}\right)-f\left(x_{0}\right)\right)+\left(f\left(x_{2}\right)-f\left(x_{1}\right)\right)+\cdots\right.\\ &\left.\quad+\left(f\left(x_{n-1}\right)-f\left(x_{n-2}\right)\right)+\left(f\left(x_{n}\right)-f\left(x_{n-1}\right)\right)\right) \\ &=\frac{b-a}{n}(f(b)-f(a)) \\ &<\epsilon . \end{aligned}\]
Hence \(f\) is integrable on \([a, b]\). \(\quad\) Q.E.D.
Let \(\varphi: \mathbb{Q} \cap[0,1] \rightarrow \mathbb{Z}^{+}\) be a one-to-one correspondence. Define \(f:[0,1] \rightarrow \mathbb{R}\) by
\[f(x)=\sum_{q \in \underset{q \leq x}{\mathbb{Q} \cap [0,1]}} \frac{1}{2^{\varphi(q)}.}\]
Then \(f\) is increasing on \([0,1],\) and hence integrable on \([0,1]\).
If \(a<b\) and \(f:[a, b] \rightarrow \mathbb{R}\) is continuous, then \(f\) is integrable on \([a, b]\).
- Proof
-
Given \(\epsilon>0,\) let
\[\gamma=\frac{\epsilon}{b-a}.\]
Since \(f\) is uniformly continuous on \([a, b],\) we may choose \(\delta>0\) such that
\[|f(x)-f(y)|<\gamma\]
whenever \(|x-y|<\delta .\) Let \(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}\) be a partition with
\[\sup \left\{\left|x_{i}-x_{i-1}\right|: i=1,2, \ldots, n\right\}<\delta .\]
If, for \(i=1,2, \dots, n\),
\[m_{i}=\inf \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\}\]
and
\[M_{i}=\sup \left\{f(x): x_{i-1} \leq x \leq x_{i}\right\},\]
then \(M_{i}-m_{i}<\gamma .\) Hence
\[\begin{aligned} U(f, P)-L(f, P) &=\sum_{i=1}^{n} M_{i}\left(x_{i}-x_{i-1}\right)-\sum_{i=1}^{n} m_{i}\left(x_{i}-x_{i-1}\right) \\ &=\sum_{i=1}^{n}\left(M_{i}-m_{i}\right)\left(x_{i}-x_{i-1}\right) \\ &<\gamma \sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right) \\ &=\gamma(b-a) \\ &=\epsilon . \end{aligned}\]
Thus \(f\) is integrable on \([a, b]\). \(\quad\) Q.E.D.
Suppose \(a<b, f:[a, b] \rightarrow \mathbb{R}\) is bounded, and \(c \in[a, b] .\) Show that if \(f\) is continuous on \([a, b] \backslash\{c\},\) then \(f\) is integrable on \([a, b]\).
Suppose \(a<b\) and \(f\) is continuous on \([a, b]\) with \(f(x) \geq 0\) for all \(x \in[a, b] .\) Show that if
\[\int_{a}^{b} f=0,\]
then \(f(x)=0\) for all \(x \in[a, b]\).
Suppose \(a<b\) and \(f\) is continuous on \([a, b] .\) For \(i=0,1, \ldots, n\), \(n \in \mathbb{Z}^{+},\) let
\[x_{i}=a+\frac{(b-a) i}{n}\]
and, for \(i=1,2, \ldots, n,\) let \(c_{i} \in\left[x_{i-1}, x_{i}\right] .\) Show that
\[\int_{a}^{b} f=\lim _{n \rightarrow \infty} \frac{b-a}{n} \sum_{i=1}^{n} f\left(c_{i}\right).\]
In the notation of Exercise \(7.3 .3,\) we call the approximation
\[\int_{a}^{b} f \approx \frac{b-a}{n} \sum_{i=1}^{n} f\left(c_{i}\right)\]
a right-hand rule approximation if \(c_{i}=x_{i},\) a left-hand rule approximation if \(c_{i}=x_{i-1},\) and a midpoint rule approximation if
\[c_{i}=\frac{x_{i-1}+x_{i}}{2}.\]
These are basic ingredients in creating numerical approximations to integrals.