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{\bf Math 222: Zero Content and Theory, optional problems} \hfill  %outlier malcom
\begin{enumerate}


%\item Multi Warm Up \\
%Reparametrize the below curve by arc length from $(1,1,0)$ in the direction of increasing $t$
%\[
%r(t)= \langle e^t,  e^{-t}, t\sqrt{2}   \rangle.
%\]
%Then compute $\mathbf{T}, \mathbf{N}, \mathbf{B}$, $k(s)$, and $\tau(s)$.


\item (Folland 4.1) Prove that if $f$ is integrable on $[a,b]$ then $|f|$ is integrable on $[a,b]$, and 
\[
\left| \int_a^b f(x) \ dx \right| \leq \int_a^b |f(x)| \ dx. 
\]
Suggestion: You can assume $f$ is bounded on $[a,b]$.  To prove that $|f|$ is integrable, show that 
\[
S_P|f| - s_P|f| \leq S_P f - s_P f.
\]
For the inequality $|\int f| \leq \int |f|$, observe that $\pm f \leq |f|$ and use Theorem 4.9(c), namely that if $f,g$ are integrable on $[a,b]$ with $f(x) \leq g(x)$ for all $x \in [a,b]$, then $\int_a^b f \leq \int_a^b g$.  You'll want to use the following lemma: 
\item[Lem. 4.5:] If $f$ is a bounded function on $[a,b]$, the following conditions are equivalent:
\begin{enumerate} 
\item $f$ is integrable on $[a,b]$.
\item For every $\epsilon > 0$ there is a partition $P$ of $[a,b]$ such that $S_P f - s_P f < \epsilon$.
\end{enumerate}
 
\item (Folland 4.1) Let $f(x)=1$ if $x$ is rational, $f(x) =0$ if $x$ is irrational.  Show that $f$ is not Riemann integrable on any interval. 

\smallskip
Fun Fact:  This characteristic/indicator function of the rationals is called the Dirichlet function, $f(x)=\chi_{\Q}(x)$. In Math 425 after defining the Lebesgue integral, you'll see $\int_{\R} \chi_{\Q}(x) = 0 $ because it is zero except on the set of rational numbers, which are countable, and hence comprise a set of zero Lebesgue measure in $\R$.  

\item[Def:] Recall that a set $Z \subset \R^n$ is said to have \textbf{zero content} if for every $\varepsilon >0$ there is $\{R_1,...,R_M\}$ such that $Z \subset \cup_1^M R_m$ and $\sum_{i=1}^M \mbox{vol}(R_m)  \leq \varepsilon$. 


\item (Folland 4.1) Let $\{ x_k\}$ be a convergent sequence in $\R$.  Show that the set $\{ x_1, x_2, ... \} $ has zero content in $\R$.

\item (Angenent) Let $f: [a,b] \to \R$ be a nondecreasing function (e.g. if $x \leq x'$ then $f(x) \leq f(x')$).  Show that the graph of $f$, i.e. the set
\[
G = \{(x,f(x)) \ | \ a \leq x \leq b \}
\]
has zero content in $\R^2$.  
Suggestion: For any (large) integer $N$, consider the rectangles
\[
R_j = [x_{j-1}, x_j] \times [ f(x_{j-1}), f(x_j)], \ \ \ 1 \leq j \leq N,
\]
where $x_j = a + \frac{j}{N}(b-a)$. There is a nice formula for the total area of the rectangles $R_1,...,R_N$.

\item (Folland 4.2) Let $f: [a,b] \to \R$ be an integrable function.
\begin{enumerate}
\item Show that the graph of $f$ in $\R^2$ has zero content.  
\item[]Suggestion: Given a partition $P$ of $[a,b]$, interpret $s_P f - S_P f$ as a sum of areas of rectangles that cover the graph of $f$.
\item Suppose $f \geq 0$, and let $S=\{ (x,y): x \in [a,b], \ 0 \leq y \leq f(x) \}$.  Show that $S$ is measurable, e.g. $S$ is bounded and and its boundary $\partial S$ has zero content in $\R^2$, and that its area (which we more precisely defined below) equals $\int_a^b f(x) dx$.  
\end{enumerate}

\item[Def:] We can precisely define \textbf{area} as follows:
\[
\mbox{area}(S) = \iint_S 1 \ dA := \iint_R \chi_S dA,
\]
where $R$ is any rectangle containing $S$.  This means, that given any bounded set $S \subset \R^2$, to compute $\iint_S 1 \ dA$, we enclose $S$ in a large rectangle $R$ and consider a partition $P$ of $R$, which produces a grid of small rectangles that cover $S$.  The lower sum for this partition is simply the sums of the areas of the small rectangles that are contained in $S$, whereas the upper sum is the sum of the areas of the small rectangles that intersect $S$.  


Taking the suprememum of the lower sums and the infimum of the upper sums yields quantities which we define to be the \textbf{inner area} and \textbf{outer area} of $S$:
\[
\underline{A}(S) = \underline{\int}_R(\chi_S) := \sup_P s_P (\chi_S) , \ \ \ \overline{A}(S) = \overline{\int}_R(\chi_S):=\inf_P S_P (\chi_S).
\]
When $\underline{A}(S) =\overline{A}(S)$, that is when the characteristic function $\chi_S$ is integrable, this common value of the inner and outer areas is defined to be the \textbf{area} of $S$. \\ \vspace{-.125in}
\begin{figure}[h]
\begin{centering}
  \includegraphics[width=.35\linewidth]{inner-outer-area.png}
  \caption{Approximations to the inner and outer areas of a region}
  \label{fig:boat1}
\end{centering}
\end{figure}

\item[Remarks:]From the next set of exercises, we can conclude that the inner and outer areas of a bounded set $S$ coincide precisely when the outer area of the boundary $\partial S$ is zero.   

But this merely means the condition that the boundary $\partial S$ should have zero content in $\R^2$.  In short, the inner and outer area of $S$ coincide precisely when $S$ is measurable.  (A set $S \subset \R^2$ is said to be \textbf{measurable} whenever $S$ is bounded and $\partial S$ has zero content in $\R^2$.). 

This is the explanation for the name \textbf{measurable}: The measurable sets in $\R^2$ are those with a well-defined area.  One analogously defines the higher dimensional area, otherwise known as volume, for measurable sets in any dimension $\R^n$.  

\item (Folland 4.2, Jones 10.A) Let $S$ be a bounded set in $\R^2$.  Show that $S$ and its interior, $S^{\mbox{\scriptsize int}}$ have the same inner area.  \\
Suggestion: For any rectangle contained in $S$, there are slightly smaller rectangles contained in $S^{\mbox{\scriptsize int}}$.

\item (Folland 4.2, Jones 10.A) Let $S$ be a bounded set in $\R^2$.  Show that $S$ and its closure $\overline{S}$ have the same outer area.\\
Suggestion: For any rectangle that intersects $\overline{S}$, there are slightly larger rectangles that intersect $S$.  

\item (Folland 4.2, Jones 10.A)
Let $S$ be a bounded set in $\R^2$.  Show that the inner area of $S$ plus the outer area of $\partial S$ equals the outer area of $S$.  \\
Suggestion: Use the preceding two exercises.
\end{enumerate}





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