\documentclass{article}
\input{18100bpreamble}
\usepackage{amssymb,amsfonts,amsmath,mathrsfs} 

\begin{document}
\pset{7}{Dan Ports}{2003/04/25}{drkp@mit.edu}

\begin{ppart}{Problem 1}
Let $f$ be a continuous function defined on $[a,b]$ such that $f \ge 0$ and $\int_a^b f(x)~dx = 0$. We show by contradiction that $f(x) = 0$ for all $x \in [a,b]$. Suppose that there exists some point $p \in (a,b)$ such that $f(p) \neq 0$. (Note that it suffices to consider only the interior $(a,b)$ because if $f(p) \neq 0$ for one of the endpoints of the interval, then $f(p)$ must also be non-zero for some interior point by continuity). Since $f \ge 0$ on the entire interval, $f(p) > 0$. We can use the continuity of $f$ to find a value $\delta > 0$ such that $f(x) > \frac{p}{2}$ for every point within $\delta$ of $p$. We then define $\delta' = \min\{\delta, p-a, b-p\}$ (this handles the cases in which $p-\delta < a$ or $p+\delta > b$). Clearly $f(x) \ge \frac{p}{2}$ on the interval $[p-\delta',p+\delta'] \subseteq [a,b]$. 
\paragraph{}
We use this $\delta'$ to define a partition $P = \{a,p-\delta',p+\delta',b\}$. We show that the lower Riemann sum $L(P,f)$ is greater than zero. Since $f \ge 0$ on $[a,b]$, the $\inf f$ over the sub-intervals $[a,p-\delta']$ and $[p+\delta']$ are non-negative, and thus they do not contribute negatively to the lower Riemann sum, so $L(P,f)$ is bounded below by the component from the $[p-\delta',p+\delta']$ interval:
\[ L(P,f) \ge 2\delta' \inf_{\{p-\delta' \leq x \leq p+\delta'\}} f(x) > 2\delta' \frac{p}{2} > 0 \]
The integral over the interval $[a,b]$ is bounded below by this:
\[ \int_a^b f(x)~dx = \underline{\int}_a^b f(x) dx \ge L(P,f) > 0 \]
We have just shown that $\int_a^b f(x)~dx$ is greater than zero, contradicting our assumption that it is zero. Thus no such $p$ can exist; $f(p) = 0$ for all $x \in [a,b]$.
\end{ppart}

\begin{ppart}{Problem 2}
Let $f$ be a monotonically decreasing function defined on $[1,\infty)$ such that $\forall x ~ f(x) \geq 0$. We will show that $\int_1^\infty f(x)~dx$ converges if and only if $\sum_{n=1}^\infty f(n)$ converges. Note that since $f$ is monotonically decreasing, $f \in \mathscr{R}$ on $[1,b]$ for any $b > 1$.
\paragraph{}
Suppose the series $\sum_{n=1}^\infty f(n)$ converges. We will show that $\int_1^\infty f(x)~dx = \lim_{n \to \infty} \int_1^n f(x)~dx$ converges. Let $n$ be an integer greater than $1$, and define a partition $P_n$ of the interval $[1,n]$ to be the integers $\{1, 2, \ldots, n\}$. Since $f$ is monotonically decreasing, $\sup_{\{i\leq x \leq i+1\}} f(x) = f(i)$ and $\inf_{\{i\leq x \leq i+1\}} f(x) = f(i+1)$. Therefore, since the length between each point in the partition is 1, the upper Riemann sum $U(P_n,f) = \sum_{i=1}^{n-1} f(i)$; and similarly $L(P_n,f) = \sum_{i=2}^n f(i)$. We use this to show that $\lim_{n \to \infty} \int_1^n f(x)~dx$ converges by the Cauchy criterion. Given some $\epsilon > 0$, we can find a $N$ such that $\forall p,q \ge N$ the $p$th and $q$th partial sums (call them $S_p$ and $S_q$) of $\sum f(n)$ are within $\epsilon$ of each other. Let $M = N+1$. Then for any $m,n > M$,
%\begin{eqnarray*}
\[S_{n-1} = U(P_n,f) \ge \int_1^n f(x)~dx \ge L(P_n,f) = S_n\]
\[S_{m-1} = U(P_m,f) \ge \int_1^m f(x)~dx \ge L(P_m,f) = S_m \]
%\end{eqnarray*}
Subtracting these two inequalities, we find
\[ S_{n-1} - S_{m-1} \ge \int_1^n f(x)~dx - \int_1^m f(x)~dx \ge S_n - S_m \]
We know $m$, $n$, $m-1$, and $n-1$ are all greater than $N$, so $\abs{S_{n-1} - S_{m-1}} \leq \epsilon$ and $\abs{S_{n} - S_{m}} \leq \epsilon$. Thus $\int_1^n f(x)~dx - \int_1^m f(x)~dx \le \epsilon$. This means that $\lim_{n \to \infty} \int_1^n f(x)~dx = \int_1^\infty f(x)~dx$ satisfies the Cauchy criterion, so it converges.
\paragraph{}
Next suppose that $\int_1^\infty f(x)~dx = \lim_{n \to \infty} \int_1^n f(x)~dx$ converges. We will show that $\sum_{n=1}^\infty f(n)$ converges. Observe that applying Rudin's theorem 6.12(c) to the integral gives $\lim_{n \to \infty} \int_1^n f(x)~dx$ \- $= \lim_{n \to \infty} \sum_{i=1}^{n-1} \int_i^{i+1} f(x)~dx$. Thus the convergence of $\int_1^\infty f(x)~dx$ is equivalent to the convergence of the series $\sum_{i=1}^\infty \int_i^{i+1} f(x)~dx$. Since $f(x)$ is monotonically decreasing, $\int_i^{i+1} f(x)~dx \ge 1\inf_{\{i\le x \le i+1\}}f(x) = f(i+1)$. Thus the series $\sum_{i=1}^\infty f(i+1)$ converges by the comparison test since it is termwise less than a series that converges. We can certainly add in the term $f(1)$ without affecting the convergence of the series, so $\sum_{i=1}^\infty f(i)$ converges.
\end{ppart}

\begin{ppart}{Problem 3}
Let $f$ and $g$ be functions such that $f \ge 0$, $g \ge 1$, and $f,g \in \mathscr{R}$ on $[a,b]$. We show that $\frac{f}{g} \in \mathscr{R}$ on $[a,b]$. To do this, we first show that $\frac{1}{g} \in \mathscr{R}$ on $[a,b]$. Since $g$ is Riemann-integrable, it is bounded. We can express this as $1 \le g \le M$ for some $M \in \mathbb{R}$. Let $\phi(x) = \frac{1}{x}$. $\phi(x)$ is clearly continuous on the domain $[1, M]$, so we can apply Rudin's theorem 6.11. This tells us that $\phi(g(x)) = \frac{1}{g(x)} \in \mathscr{R}$ on $[a,b]$. We next apply theorem 6.13(a) and find that $f \frac{1}{g} = \frac {f}{g} \in \mathscr{R}$ on $[a,b]$.

\end{ppart}

\begin{ppart}{Problem 4}
Let $f$ be a bounded function on $[a,b]$ that is continuous except at a countable set $E$ of points. We show that $f \in \mathscr{R}$ on $[a,b]$.
\paragraph{}
Consider the function $r(x) = \lim_{k\to \infty} (\sup f - \inf f)$ on the interval $(x-\frac{1}{k},x+\frac{1}{k})$. Define a collection of sets $B_n = \{x \colon r(x) < \frac{1}{n}\}$ for $n \in \mathbb{J}$. Each set $B_n$ is open: consider any $x \in B_n$. Then $r(x) = m$ for some $m < \frac{1}{n}$. That is, $\lim_{k\to \infty} (\sup f - \inf f)$ on $(x-\frac{1}{k},x+\frac{1}{k})$ is $m$, which means we can find an interval $(x-\frac{1}{k},x+\frac{1}{k})$ where $\sup f - \inf f$ is arbitrarily close to $m$. Since $m < \frac{1}{n}$, we can certainly find such an interval where $(\sup f - \inf f) < \frac{1}{n}$. Every point $p$ in this interval, which is a neighborhood of $x$, will thus have $r(p) < \frac{1}{n}$, implying $p \in B_n$. Thus every point $x$ of $B_n$ is an interior point, so every $B_n$ is open. Hence, $B_n^c$ is closed. There are a countable number of such sets, so $\bigcup_{n=1}^\infty B_n^c$ is closed.
\paragraph{}
By definition of continuity, if $r(x) = 0$, $f(x)$ is continuous at $x$ because the sup and inf are equal, and if $r(x) \neq 0$, $f(x)$ cannot be continuous at $x$. By definition of $B_n$, $B_n^c$ contains the points $x$ such that $r(x) \ge \frac{1}{n}$, so their union will contain all points where $r(x) > 0$. This is precisely the set $E$ of all points at which $f$ is discontinuous. Since $E$ is countable, each set $B_n^c \subseteq E$ must be finite or countable.
\paragraph{}
For any $m$, consider the set $B_m^c$. This set contains all $x$ for which $r(x) \ge \frac{1}{m}$. It is closed (and bounded, therefore compact), and also countable. We can therefore call its elements $\{x_1, x_2, \ldots\}$. For each element $x_i$, we consider the neighborhood (interval) of radius $\frac{2^{-i}}{m}$ around $x_i$. These are all open sets, and their union certainly contains $B_m^c$, which is compact, so there exists a finite subcover that contains $B_m^c$. This subcover is a finite set of intervals of length $\frac{2^{-i}}{m}$; their total length is strictly less than $\sum_{i=1}^\infty \frac{2^{-i}}{m} = \frac{1}{m}$. We will designate this subcover as the collection of open sets $\{C_1, C_2, \ldots, C_k\}$, and let $C = \bigcup_{i=1}^k C_i$. $C$ contains all $x$ for which $r(x) \ge \frac{1}{m}$.
\paragraph{}
Now we consider the set (call it $F$) that consists of the interval $[a,b]$ with $C$ removed. Every $x$ in this set has $r(x) < \frac{1}{m}$. By definition of $r(x)$, there exist intervals $(x-\frac{1}{k},x+\frac{1}{k})$ where $\sup f - \inf f$ is arbitrarily close to $r(x)$; we can thus find such an interval (call it $N(x)$) where $\sup f - \inf f < \frac{1}{m}$ since $r(x) < \frac{1}{m}$. These intervals are open sets, so the union $\bigcup_{x \in F} N(x)$ is an open cover of $F$. Since $C$ is open and $F = [a,b] \cap C^c$, $F$ is compact. So the open cover can be reduced to a finite subcover $D = N(x_1) ~\cup~ N(x_2) ~\cup~ \cdots ~\cup~ N(x_k)$. Since this still covers $F$, we can reduce each open interval $N(x_i)$ to a closed subinterval $N'(x_i)$ such that every point in $F$ is in exactly one $N'(x_i)$. The resulting union $D' = \bigcup_{n=1}^\infty N'(x_i)$ contains every point in $F$, and has total length at most $b-a$.
\paragraph{}
Note that $C \cup D' = [a,b]$. This arrangement of sets $C$ and $D'$ allows us to make the upper and lower Riemann sums arbitrarily close. Letting $P$ be the partition defined by the endpoints of the intervals in $C$ and $D'$, and $M = \sup \abs{f(x)}$, we show:
\[U(P,f) - L(P,f) < \frac{2M}{m} + \frac{b-a}{m}\]
The first term of the inequality comes from the sets in $C$. The sup and inf on these sets can differ by as much as $2M$, but their length is less than $\frac{1}{m}$. The second term comes from the sets in $D'$. The sup and inf on these sets can only differ by less than $\frac{1}{m}$ (since they only contain points where $r(x) < \frac{1}{m}$), and the total length of these sets is at most $b-a$. Since $m$ was arbitrary, given any $\epsilon > 0$, we can choose $m = \frac{2M + b - a}{\epsilon}$, which makes the upper and lower Riemann sums within $\epsilon$ of each other. Thus, $f \in \mathscr{R}$ on $[a,b]$.
\end{ppart}
\end{document}