(32.2) Let the function \(f\) be defined as,
\[f(x) = \left\{ \begin{array}{ll} x\;\; & x \in \mathbb Q \\ 0\;\; & x \in \mathbb R \setminus \mathbb Q \end{array} \right.\]Compute the upper and lower Darboux integrals for \(f\) on the interval \([0,b]\).
Hint. Here’s how you can show that \(U(f) \ge b^2/2\): For any partition \(Q\), if \(P_n = \{\cdots \frac{kb}n \cdots\}\), explain why
\[U(f, Q \cup P_n) \ge \sum_{k=1}^{n}{ \frac{(k-1)b}{n} \left(\frac bn\right) }.\]Is \(f\) integrable on the interval \([0, b]\)?
For any interval \(I\), by density of the irrationals, we have that \(m(f, I) = 0\). So, for any partition \(P\) of \([0, b]\), we have that \(L(f,P) = 0\), so we conclude that \(L(f) = 0\).
Consider now the partition \(P_n = \left\{ \cdots, \frac{kb}n, \cdots \right\}\). We have, by density of the rationals and the definition of \(f\) that
\[U(f, P_n) = \sum_{k=1}^n {\frac{kb}{n}\left( \frac bn \right)} = \frac {b^2}{n^2} \sum_{k=1}^n k = \frac{b^2 n(n+1)}{2n^2}\]So, \(\lim_{n\to\infty} U(f, P_n) = \frac {b^2}{2}\), and as \(U(f)\) is the infimum over all \(U(f, P)\), we have proven that \(U(f) \le \frac {b^2}2\).
To prove the other inequality, let \(Q\) be any partition of \([0, b]\). Observe that
\[U(f, Q) \ge U(f, Q \cup P_n) \ge \sum_{k=1}^{n} \frac{(k-1)b}{n} \left( \frac bn \right) = \frac{b^2 (n-1)n}{2n^2}\]This inequality holds for all \(n\), so \(U(f, Q) \ge \frac{b^2}2\) for all partitions \(Q\) of \([0, b]\). From this we notice that \(U(f) \ge \frac{b^2}2\) as it is a greatest lower bound. Together now, we conclude that \(U(f) = \frac{b^2}2\).
No, as in part (a) we have proved that \(U(f) \ne L(f)\).
Let \(f\) be a bounded function on \([a, b]\). Suppose that \(f^2\) is integrable. Does it follow that \(f\) is also integrable? If so, prove it. If not, provide a counterexample.
No. Let \(f(x) = 1\) for \(x \in \mathbb Q\) and \(f(x) = -1\) for irrational \(x\). Then \(f^2\) is integrable but \(f\) is not by a calculation similar to Example 22.2.
(32.6) Let \(f\) be a bounded function on \([a, b]\). Suppose there exist sequences \((U_n)\) and \((L_n)\) of upper and lower Darboux sums for \(f\) such that \(\lim(U_n - L_n) = 0\). Show \(f\) is integrable on \([a,b]\) and that \(\int_a^b f = \lim U_n = \lim L_n\).
Proof. Notice first that the sequences \(U_n\) and \(L_n\) are defined by two sequences of partitions \((P_n)\) and \((Q_n)\) for which \(U_n = U(f, P_n)\) and \(L_n = L(f, Q_n)\). In particular, we have that \(U_n \ge L_n\) as they are upper and lower Darboux sums.
Let \(\epsilon > 0\). Then there exists \(N\) so that for \(n > N\) we know that \(U_n - L_n < \epsilon\). Notice that
\(\epsilon > U_n - L_n = U(f, P_n) - L(f, Q_n) \ge U(f, P_n\cup Q_n) - L(f, P_n \cup Q_n)\).
Take any \(n > N\) and let \(P = P_n \cup Q_n\). We have just proved that \(P\) is a partition for which \(U(f, P) - L(f, P) < \epsilon\). In other words, we have proved that \(f\) is integrable.
Notice that \(L_n \le U(f) \le U_n\) for all \(n\), so \(0 \le U(f) - L_n \le U_n - L_n\). In particular, when \(n > N\), we see that \(U(f) - L_n \le U_n - L_n < \epsilon\). So \(U(f) = \lim L_n\); in particular, \(L_n\) converges. As \(\lim U_n - L_n = 0\), we conclude furthermore that \(\lim L_n = \lim U_n = U(f) = \int_a^b f\).
(32.7) Let \(f\) be integrable on \([a, b]\) and suppose \(g\) is a function on \([a,b]\) so that \(g(x) = f(x)\) except perhaps at finitely many \(x\) in \([a, b]\).
Show that \(g\) is integrable on \([a, b]\) and that \(\int_a^b f = \int_a^b g\). Hint: First reduce to the case where \(f(x) = 0\) for all \(x\).
Proof. Suppose \(f(x) = 0\) and that \(g(x) = 0\) at almost all of \([a, b]\), except for when \(x = c\) we have \(g(x) = \gamma\). Suppose first that \(\gamma > 0\).
Let \(\epsilon > 0\). If \(c = a\), consider the partition \(P = \{ a, a+\epsilon/(2\gamma), b \}\). If \(c = b\), consider the partition \(P = \{a, b-\epsilon/(2\gamma), b \}\). Otherwise, consider the partition \(P = \{a, c-\epsilon/(4\gamma), c+\epsilon/(4\gamma)\}\). In all cases, observe that \(L(f, P) = 0\) and that \(U(f, P) = \epsilon/2\). In particular,
\[U(f, P) - L(f, P) = \gamma \epsilon/(2\gamma) = \epsilon/2 < \epsilon,\]So \(g\) is integrable and \(\int_a^b g = L(f) = 0 = \int_a^b f\).
Suppose instead now that \(\gamma < 0\); but by considering the function \(-g\), we have just proved that \(\int_a^b (-g) = 0\), and we may use linearity of the integral to realize that \(\int_a^b f = 0 = - \int_a^b (-g) = \int_a^b -(-g) = \int_a^b g\) in this case as well.
Now let \(f\) be any function integrable on \([a, b]\), and suppose that \(g(x) = f(x)\) except at some point \(x = c\). Considering the function \((g - f)\) which is 0 at all points except for \(x = c\), realize that we have shown above that \(0 = \int_a^b (g-f)\), and by linearity of the integral,
\[\int_a^b f = \int_a^b (g-f) + \int_a^b f = \int_a^b (g-f)+f = \int_a^b g.\]We have now shown completely the case in which \(g\) differs by integrable \(f\) at precisely one point. But what about finitely many points? We may use induction! Suppose that \(g_k(x) = f(x)\) at all but \(k\) points \(x_1, \cdots x_k\), and take as an inductive hypothesis that \(\int_a^b g_k = \int_a^b f\). Then notice that if \(g_{k+1}(x) = f(x)\) at all but the \(k+1\) points \(x_1, \cdots, x_k, x_{k+1}\) we have that \(g_{k+1}(x) = g_k(x)\) actually at all but one point \(x_{k+1}\). But we have shown above that
\[\int_a^b g_{k+1} = \int_a^b g_k = \int_a^b f.\]So, we have proven the case for any finite number of differing points. \(\Box\)