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\title{Homework assignment 7}
\author{Anthony Beal}
\date{}
\maketitle

\begin{enumerate}
\item Let $Q$ be the open rectangle $(0,1)^2$ in $\R^2$, let $A$ be the subset of $Q$ with rational coordinates, and let $B$ be a collection of open rectangles in $Q$ which contains $A$.
\begin{enumerate}
\item Prove that $A$ has measure zero and $Q-A$ does not have measure zero. \\
$\boldsymbol{Solution:}$ The set $A$ is a subset of $\mathbb{Q}^2$. Since $\mathbb{Q}^2$ is countable, $A$ must have has measure zero. If $Q-A$ also had measure zero, then $A \cup (Q-A) = Q$ would have measure zero, which is impossible. Thus, $Q-A$ does not have measure zero.
\item Prove that the closure $\overline{B}$ of $B$ is the closed rectangle $[0,1]^2$.\\
$\boldsymbol{Solution:}$ Let $x$ be any point in $[0,1]^2$. For any $\varepsilon > 0, $ the ball $B_\varepsilon(x)$ intersects $B$ at some $b \neq x$. Thus, $x$ is a limit point of $B$ and must therefore be in $\overline{B}.$ Points not contained in $[0,1]^2$ cannot be in $B$ and are not limit points of $B$, and so they cannot be in $\overline{B}.$ We can therefore conclude that $\overline{B} = [0,1]^2.$
\item By part (a), the collection $B$ can be chosen so that its total volume is as small as desired. Prove that if the total volume of $B$ is sufficiently small, then the boundary $\partial B$ does not have measure zero. (Try using the result of part (b).)\\
$\boldsymbol{Solution:}$ Since $\overline{B} = \text{Int}(B) \cup \partial B$ does not have measure zero, there exists an $\overline{\varepsilon} > 0$ such that the sum of the volumes of any collection of rectangles covering $\overline{B}$ is greater than $\overline{\varepsilon}$. We've previously showed that we can choose any collection of rectangles covering $\text{Int}(B)$ to have the sum of their volumes be less than $\frac{\overline{\varepsilon}}{2}.$ Then, $\partial B$ cannot be covered by any collection of rectangles whose volumes sum to be less than $\frac{\overline{\varepsilon}}{2}$. In other words, $\partial B$ fails to have measure zero whenever the total volume of $B$ is made sufficiently small.

\item By part (c), the collection $B$ can be chosen so that $\partial B$ does not have measure zero. In this case, prove that $\int_B 1$ does not exist as an ordinary integral, but that $\int_B 1$ does exist as an improper integral, and its value is strictly positive.\\
$\boldsymbol{Solution:}$ Choose a collection of $B$ for which $\partial B$ does not have measure zero. To show that $\int_B 1$ does not exist as an ordinary integral, we will let $1_B$ denote the indicator function of $B$ and show that $\int_B 1_B$ does not exist (i.e. that $B$ is not rectifiable). Since $1_B$ is discontinuous on $\partial{B}$, and $\partial{B}$ does not have measure zero, $B$ is not rectifiable, meaning $\int_B 1$ does not exist. Next, to prove the existence of $\partial B$ as an improper integral, we will utilize theorem 15.2. Let $C_N$ be any sequence of compact rectifiable subsets of $B$ whose union is $B$ such that $C_N \subset \text{Int} C_{N+1}$ for each $N$. Since each $C_N$ is contained in $Q$, $$\int_{C_N} |1| \leq Vol(Q).$$ Therefore, by theorem 15.2, $\int_B 1$ as an improper integral exists.
\end{enumerate}

\item Suppose $A$ is an open set, $f$ and $g$ are continuous functions from $A$ to $\R$, $g(x) \leq f(x)$ everywhere, and $\int_A f$ exists (as an improper integral).
\begin{enumerate}
\item If $\int_A g$ also exists (as an improper integral), prove that $\int_A g \leq \int_A f$. (As the book mentions, and as we mentioned in class, this follows from the definition of the improper integral and the similar fact for ordinary integrals; write out the details.) \\
$\boldsymbol{Solution:}$ Since $g(x) \leq f(x)$, it follows that $g_{+} \leq f_{+}$ and $-g_{-} \leq -f_{-}$. So, $\int_A g_{+} - \int_A g_{-} \leq \int_A f_{+} - \int_A f_{-} \implies \int_A g \leq \int_A f$.
\item If $|g(x)| \leq f(x)$ everywhere, prove that $\int_A g$ exists (as an improper integral).\\
$\boldsymbol{Solution:}$ Let $C_N$ be a sequence of compact rectifiable sets covering A such that $C_N \subset C_{N+1}$ for all $N \in \mathbb{N}$. Since $\int_A f$ exists as an improper integral, there exists an $M > 0$ such that $\int_{C_N} |f| < M$ for all $N$. By monotonicity, $\int_{C_N} |g| \leq \int_{C_N} |f| < M$. In other words, $\int_{C_N} |g|$ is bounded above by $M$, and thus we can conclude $\int_{C_N} g$ exists as an improper integral.
\item Find an example showing that the absolute value condition in part (b) is necessary. That is, find an open set $A$ and continuous functions $f,g$ from $A$ to $\R$ such that $g(x) \leq f(x)$ everywhere, $\int_A f$ exists (as an improper integral), and $\int_A g$ does not exist (as an improper integral).\\
$\boldsymbol{Solution:}$ Let $A = (0,1)$, $g(x) = \frac{-1}{x}, $ and $f(x) = e^\frac{-1}{x}$.

\end{enumerate}

\item Suppose $A$ is an open set, $f \colon A \to \R$ is continuous, and $\int_A f$ exists (as an ordinary integral). Prove that $\int_A f$ exists (as an improper integral), and that the two integrals are equal.\\
$\boldsymbol{Solution:}$ Let $C_N$ be a sequence of compact rectifiable sets covering A such that $C_N \subset C_{N+1}$ for all $N \in \mathbb{N}$. Assume that $\int_A f$ does not exist as an improper integral (i.e. that $\int_{C_N} |f|$ is an unbounded sequence). Thus, we can pick some $C_k$ such that $\int_{A} |f| < \int_{C_k} |f| $, which is obviously nonsense. Therefore, $\int_A f$ must exist as an improper integral.\\



\item Define $f \colon \R \to \R$ by $f(x)=x$.
\begin{enumerate}
\item Prove that $\int_\R f$ does not exist (as an improper integral).\\
$\boldsymbol{Solution:}$ Let $C_N = [-N,2N].$ Then, the sequence $\int_{C_N} |x|$ is unbounded, and so the improper integral does not exist.
\item For any real number $\lambda$, find a sequence $C_i$ of compact rectifiable subsets of $\R$ such that the union of the $C_i$ is equal to $\R$, each $C_i$ is contained in the interior of $C_{i+1}$, and $\displaystyle \lim_{i \to \infty} \int_{C_i} f = \lambda$.\\
$\boldsymbol{Solution:}$ Let $C_n = [-\sqrt{2\lambda (n-1)}, \sqrt{2\lambda n}]$ for $n \geq 1$.
\end{enumerate}

\item Let $A = \{(x,y) \in \R^2 : x>1, \, y>1\}$ and $B=\{(x,y) \in \R^2 : 0<x<1, \, 0<y<1\}$. Define $f \colon A \cup B \to \R$ by $f(x,y)=(xy)^{-1/2}.$
\begin{enumerate}
\item Calculate the improper integral $\int_A f$ or show it does not exist.\\
$\boldsymbol{Solution:}$ Let $C_n = [1,n]^2$ and observe that the sequence $\int_{1}^n \int_{1}^n \frac{1}{\sqrt{xy}} = (2\sqrt{n}-2)^2$ is unbounded. Therefore $\int_A f$ does not exist as an improper integral.

\item Calculate the improper integral $\int_B f$ or show it does not exist.\\
$\boldsymbol{Solution:}$

$$\lim_{n \to \infty} \int_{\frac{1}{n}}^{\frac{n}{n+1}} \int_{\frac{1}{n}}^{\frac{n}{n+1}} \frac{1}{\sqrt{xy}} = \lim_{n \to \infty} (2\sqrt{\frac{n}{n+1}} - 2\sqrt{\frac{1}{n}})^2 = 4.$$
\end{enumerate}
\emph{Note:} examples 2 and 3 of section 15 may be worth reading, as they describe how to calculate improper integrals like these.

\item Let $A = \{(x,y) \in \R^2 : 0<x<y<2x\}$ and $B=\{(x,y) \in \R^2 : x>0, \, 0<x^2<y<2x^2\}$. Define $f \colon A \cup B \to \R$ by $f(x,y)=(y+1)^{-2}$.
\begin{enumerate}
\item Show that the improper integral $\int_A f$ does not exist.
\item Calculate the improper integral $\int_B f$.
\end{enumerate}

\end{enumerate}
\end{document}