7.4: Properties of Integrals
- Page ID
- 22681
If \(f: D \rightarrow \mathbb{R}\) and \(g: D \rightarrow \mathbb{R},\) then
\[\sup \{f(x)+g(x): x \in D\} \leq \sup \{f(x): x \in D\}+\sup \{g(x): x \in D\}\]
and
\[\inf \{f(x)+g(x): x \in D\} \geq \inf \{f(x): x \in D\}+\inf \{g(x): x \in D\}\]
Prove the previous proposition.
Find examples for which the inequalities in the previous proposition are strict.
Suppose \(f\) and \(g\) are both integrable on \([a, b] .\) Then \(f+g\) is integrable on \([a, b]\) and
\[\int_{a}^{b}(f+g)=\int_{a}^{b} f+\int_{a}^{b} g.\]
- Proof
-
Given \(\epsilon>0,\) let \(P_{1}\) and \(P_{2}\) be partitions of \([a, b]\) with
\[U\left(f, P_{1}\right)-L\left(f, P_{1}\right)<\frac{\epsilon}{2}\]
and
\[U\left(g, P_{2}\right)-L\left(g, P_{2}\right)<\frac{\epsilon}{2}.\]
Let \(P=P_{1} \cup P_{2} .\) By the previous proposition,
\[U(f+g, P) \leq U(f, P)+U(g, P)\]
and
\[L(f+g, P) \geq L(f, P)+L(g, P).\]
Hence
\[\begin{aligned} U(f+g, P)-L(f+g, P) & \leq(U(f, P)+U(g, P))-(L(f, P)+L(g, P)) \\ &=(U(f, P)-L(f, P))+(U(g, P)-L(g, P)) \\ & \leq\left(U\left(f, P_{1}\right)-L\left(f, P_{1}\right)\right)+\left(U\left(g, P_{2}\right)-L(g, 2 P)\right) \\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon . \end{aligned}\]
Hence \(f+g\) is integrable on \([a, b]\).
Moreover,
\[\begin{aligned} \int_{a}^{b}(f+g) & \leq U(f+g, P) \\ & \leq U(f, P)+U(g, P) \\ & \leq\left(\int_{a}^{b} f+\frac{\epsilon}{2}\right)+\left(\int_{a}^{b} g+\frac{\epsilon}{2}\right) \\ &=\int_{a}^{b} f+\int_{a}^{b} g+\epsilon \end{aligned}\]
and
\[\begin{aligned} \int_{a}^{b}(f+g) & \geq L(f+g, P) \\ & \geq L(f, P)+L(g, P) \\ & \geq\left(\int_{a}^{b} f-\frac{\epsilon}{2}\right)+\left(\int_{a}^{b} g-\frac{\epsilon}{2}\right) \\ &=\int_{a}^{b} f+\int_{a}^{b} g-\epsilon . \end{aligned}\]
Since \(\epsilon>0\) was arbitrary, it follows that
\[\int_{a}^{b}(f+g)=\int_{a}^{b} f+\int_{a}^{b} g.\]
Q.E.D.
Suppose \(a<b\) and \(f:[a, b] \rightarrow \mathbb{R}\) and \(g:[a, b] \rightarrow \mathbb{R}\) are both bounded. Show that
\[\overline{\int_{a}^{b}}(f+g) \leq \overline{\int_{a}^{b}} f+\overline{\int_{a}^{b}} g.\]
Find an example for which the inequality is strict.
Find an example to show that \(f+g\) may be integrable on \([a, b]\) even though neither \(f\) nor \(g\) is integrable on \([a, b]\).
If \(f\) is integrable on \([a, b]\) and \(\alpha \in \mathbb{R},\) then \(\alpha f\) is integrable on \([a, b]\) and
\[\int_{a}^{b} \alpha f=\alpha \int_{a}^{b} f.\]
Prove the previous proposition.
Suppose \(a<b, f:[a, b] \rightarrow \mathbb{R}\) is bounded, and \(c \in(a, b)\). Then \(f\) is integrable on \([a, b]\) if and only if \(f\) is integrable on both \([a, c]\) and \([c, b] .\)
- Proof
-
Suppose \(f\) is integrable on \([a, b] .\) Given \(\epsilon>0,\) let \(Q\) be a partition of \([a, b]\) such that
\[U(f, Q)-L(f, Q)<\epsilon .\]
Let \(P=Q \cup\{c\}, P_{1}=P \cap[a, c],\) and \(P_{2}=P \cap[c, b] .\) Then
\[\begin{aligned}\left(U\left(f, P_{1}\right)-L\left(f, P_{1}\right)\right)+\left(U\left(f, P_{2}\right)-L\left(f, P_{2}\right)\right) &=\left(U\left(f, P_{1}\right)+U\left(f, P_{2}\right)\right) \\ &-\left(L\left(f, P_{1}\right)+L\left(f, P_{2}\right)\right) \\ &=U(f, P)-L(f, P) \\ & \leq U(f, Q)-L(f, Q) \\ &<\epsilon . \end{aligned}\]
Thus we must have both
\[U\left(f, P_{1}\right)-L\left(f, P_{1}\right)<\epsilon\]
and
\[U\left(f, P_{2}\right)-L\left(f, P_{2}\right)<\epsilon.\]
Hence \(f\) is integrable on both \([a, c]\) and \([c, b]\).
Now suppose \(f\) is integrable on both \([a, c]\) and \([c, b] .\) Given \(\epsilon>0,\) let \(P_{1}\) and \(P_{2}\) be partitions of \([a, c]\) and \([c, b],\) respectively, such that
\[U\left(f, P_{1}\right)-L\left(f, P_{1}\right)<\frac{\epsilon}{2}\]
and
\[U\left(f, P_{2}\right)-L\left(f, P_{2}\right)<\frac{\epsilon}{2}.\]
Let \(P=P_{1} \cup P_{2}\). Then \(P\) is a partition of \([a, b]\) and
\[\begin{aligned} U(f, P)-L(f, P) &=\left(U\left(f, P_{1}\right)+U\left(f, P_{2}\right)\right)-\left(L\left(f, P_{1}\right)+L\left(f, P_{2}\right)\right) \\ &=\left(U\left(f, P_{1}\right)-L\left(f, P_{1}\right)\right)+\left(U\left(f, P_{2}\right)-L\left(f, P_{2}\right)\right) \\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2} \\ &=\epsilon . \end{aligned}\]
Thus \(f\) is integrable on \([a, b]\). \(\quad\) Q.E.D.
Suppose \(f\) is integrable on \([a, b]\) and \(c \in(a, b) .\) Then
\[\int_{a}^{b} f=\int_{a}^{c} f+\int_{c}^{b} f.\]
- Proof
-
If \(P\) and \(Q\) are partitions of \([a, c]\) and \([c, b],\) respectively, then
\[U(f, P)+U(f, Q)=U(f, P \cup Q) \geq \int_{a}^{b} f.\]
Thus
\[U(f, P) \geq \int_{a}^{b} f-U(f, Q),\]
so
\[\int_{a}^{c} f= \overline{\int_{a}^{c}} f \geq \int_{a}^{b} f-U(f, Q) .\]
Hence
\[U(f, Q) \geq \int_{a}^{b} f-\int_{a}^{c} f,\]
so
\[\int_{c}^{b} f= \overline{\int_{c}^{b}} f \geq \int_{a}^{b} f-\int_{a}^{c} f.\]
Thus
\[\int_{a}^{c} f+\int_{c}^{b} f \geq \int_{a}^{b} f.\]
Similarly, if \(P\) and \(Q\) are partitions of \([a, c]\) and \([c, b],\) respectively, then
\[L(f, P)+L(f, Q)=L(f, P \cup Q) \leq \int_{a}^{b} f.\]
Thus
\[L(f, P) \leq \int_{a}^{b} f-L(f, Q),\]
so
\[\int_{a}^{c} f= \underline{\int_{a}^{c}} f \leq \int_{a}^{b} f-L(f, Q).\]
Hence
\[L(f, Q) \leq \int_{a}^{b} f-\int_{a}^{c} f,\]
so
\[\int_{c}^{b} f= \underline{\int_{c}^{b}} f \leq \int_{a}^{b} f-\int_{a}^{c} f.\]
Thus
\[\int_{a}^{c} f+\int_{c}^{b} f \leq \int_{a}^{b} f.\]
Hence
\[\int_{a}^{c} f+\int_{c}^{b} f=\int_{a}^{b} f.\]
Q.E.D.
Suppose \(f:[a, b] \rightarrow \mathbb{R}\) is bounded and \(B\) is a finite subset of \((a, b) .\) Show that if \(f\) is continuous on \([a, b] \backslash B,\) then \(f\) is integrable on \([a, b]\).
If \(f\) is integrable on \([a, b]\) with \(f(x) \geq 0\) for all \(x \in[a, b]\), then
\[\int_{a}^{b} f \geq 0.\]
- Proof
-
The result follows from the fact that \(L(f, P) \geq 0\) for any partition \(P\) of \([a, b]\). \(\quad\) Q.E.D.
Suppose \(f\) and \(g\) are both integrable on \([a, b] .\) If, for all \(x \in[a, b], f(x) \leq g(x),\) then
\[\int_{a}^{b} f \leq \int_{a}^{b} g.\]
- Proof
-
Since \(g(x)-f(x) \geq 0\) for all \(x \in[a, b],\) we have, using Propositions 7.4.2, 7.4.3, and 7.4.6,
\[\int_{a}^{b} g-\int_{a}^{b} f=\int_{a}^{b}(g-f) \geq 0.\]
Q.E.D.
Suppose \(f\) is integrable on \([a, b], m \in \mathbb{R}, M \in \mathbb{R},\) and \(m \leq f(x) \leq M\) for all \(x \in[a, b] .\) Then
\[m(b-a) \leq \int_{a}^{b} f \leq M(b-a).\]
- Proof
-
It follows from the previous proposition that
\[m(b-a)=\int_{a}^{b} m d x \leq \int_{a}^{b} f(x) d x \leq \int_{a}^{b} M d x=M(b-a).\]
Q.E.D.
Show that
\[1 \leq \int_{-1}^{1} \frac{1}{1+x^{2}} d x \leq 2.\]
Suppose \(f\) is continuous on \([0,1],\) differentiable on \((0,1)\), \(f(0)=0,\) and \(\left|f^{\prime}(x)\right| \leq 1\) for all \(x \in(0,1) .\) Show that
\[-\frac{1}{2} \leq \int_{0}^{1} f \leq \frac{1}{2}.\]
Suppose \(f\) is integrable on \([a, b]\) and define \(F:(a, b) \rightarrow \mathbb{R}\) by
\[F(x)=\int_{a}^{x} f.\]
Show that there exists \(\alpha \in \mathbb{R}\) such that for any \(x, y \in(a, b)\) with \(x<y\),
\[|F(y)-F(x)| \leq \alpha(y-x).\]
Suppose \(g\) is integrable on \([a, b], g([a, b]) \subset[c, d],\) and \(f:[c, d] \rightarrow \mathbb{R}\) is continuous. If \(h=f \circ g,\) then \(h\) is integrable on \([a, b]\).
- Proof
-
Let \(\epsilon>0\) be given. Let
\[K>\sup \{f(x): x \in[c, d]\}-\inf \{f(x): x \in[c, d]\}\]
and choose \(\delta>0\) so that \(\delta<\epsilon\) and
\[|f(x)-f(y)|<\frac{\epsilon}{2(b-a)}\]
whenever \(|x-y|<\delta .\) Let \(P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}\) be a partition of \([a, b]\) such that
\[U(g, P)-L(g, P)<\frac{\delta^{2}}{2 K}.\]
For \(i=1,2, \dots, n,\) let
\[m_{i}=\inf \left\{g(x): x_{i-1} \leq x \leq x_{i}\right\},\]
\[M_{i}=\sup \left\{g(x): x_{i-1} \leq x \leq x_{i}\right\},\]
\[w_{i}=\inf \left\{h(x): x_{i-1} \leq x \leq x_{i}\right\},\]
and
\[W_{i}=\sup \left\{h(x): x_{i-1} \leq x \leq x_{i}\right\}.\]
Finally, let
\[I=\left\{i: i \in \mathbb{Z}, 1 \leq i \leq n, M_{i}-m_{i}<\delta\right\}\]
and
\[J=\left\{i: i \in \mathbb{Z}, 1 \leq i \leq n, M_{i}-m_{i} \geq \delta\right\}.\]
Note that
\[\begin{aligned} \delta \sum_{i \in J}\left(x_{i}-x_{i-1}\right) & \leq \sum_{i \in J}\left(M_{i}-m_{i}\right)\left(x_{i}-x_{i-1}\right) \\ & \leq \sum_{i=1}^{n}\left(M_{i}-m_{i}\right)\left(x_{i}-x_{i-1}\right) \\ &<\frac{\delta^{2}}{2 K}, \end{aligned}\]
from which it follows that
\[\sum_{i \in J}\left(x_{i}-x_{i-1}\right)<\frac{\delta}{2 K}.\]
Then
\[\begin{aligned} U(h, P)-L(h, P) &=\sum_{i \in I}\left(W_{i}-w_{i}\right)\left(x_{i}-x_{i-1}\right)+\sum_{i \in J}\left(W_{i}-w_{i}\right)\left(x_{i}-x_{i-1}\right) \\ &<\frac{\epsilon}{2(b-a)} \sum_{i \in I}\left(x_{i}-x_{i-1}\right)+K \sum_{i \in J}\left(x_{i}-x_{i-1}\right) \\ &<\frac{\epsilon}{2}+\frac{\delta}{2} \\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2} \\ &=\epsilon . \end{aligned}\]
Thus \(h\) is integrable on \([a, b]\). \(\quad\) Q.E.D.
Suppose \(f\) and \(g\) are both integrable on \([a, b] .\) Then \(f g\) is integrable on \([a, b]\).
- Proof
-
Since \(f\) and \(g\) are both integrable, both \(f+g\) and \(f-g\) are integrable. Hence, by the previous proposition, both \((f+g)^{2}\) and \((f-g)^{2}\) are integrable. Thus
\[\left.\frac{1}{4}\left((f+g)^{2}-(f-g)^{2}\right)\right)=f g\]
is integrable on \([a, b]\). \(\quad\) Q.E.D.
Suppose \(f\) is integrable on \([a, b] .\) Then \(|f|\) is integrable on \([a, b]\) and
\[\left|\int_{a}^{b} f\right| \leq \int_{a}^{b}|f|.\]
- Proof
-
The integrability of \(|f|\) follows from the integrability of \(f,\) the continuity of \(g(x)=|x|,\) and Proposition \(7.4 .9 .\) For the inequality, note that
\[-|f(x)| \leq f(x) \leq|f(x)|\]
for all \(x \in[a, b] .\) Hence
\[-\int_{a}^{b}|f| \leq \int_{a}^{b} f \leq \int_{a}^{b}|f|, \]
from which the result follows. \(\quad\) Q.E.D.
Either prove the following statement or show it is false by finding a counterexample: If \(f:[0,1] \rightarrow \mathbb{R}\) is bounded and \(f^{2}\) is integrable on \([0,1],\) then \(f\) is integrable on \([0,1] .\)
7.4.1 Extended definitions
If \(f\) integrable on \([a, b],\) then we define
\[\int_{b}^{a} f=-\int_{a}^{b} f.\]
Moreover, if \(f\) is a function defined at a point \(a \in \mathbb{R},\) we define
\[\int_{a}^{a} f=0.\]
Suppose \(f\) is integrable on a closed interval containing the
points \(a, b,\) and \(c .\) Show that
\[\int_{a}^{b} f=\int_{a}^{c} f+\int_{c}^{b} f.\]