# Harrison Chapman

## Math 317: Homework 9

Due: Friday, April 19, 2019
1. (32.2) Let the function $f$ be defined as,

1. 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

2. Is $f$ integrable on the interval $[0, b]$?

1. 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

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

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 $U(f) = \frac{b^2}2$.

2. No, as in part (a) we have proved that $U(f) \ne L(f)$.

2. 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.

3. (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 $% $. 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 $% $. 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 $% $. So $U(f) = \lim L_n$; in particular, $L_n$ converges. As $\lim U_n - L_n = 0$, we conclude furthermore that $\lim L_n = \lim U_n = U(f) = \int_a^b f$.

4. (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,

So $g$ is integrable and $\int_a^b g = L(f) = 0 = \int_a^b f$.

Suppose instead now that $% $; 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,

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

So, we have proven the case for any finite number of differing points. $\Box$