





%\setcounter{section}{1}
\section{Introduction to the Riemann integral}

\begin{enumerate}[1.]


\item {\it 
Let $f(x,y)=x+y$, and let $D=\{(x,y):0\leq x\leq 1,0\leq y\leq 1\}$. Prove that for any $\epsilon>0$, there is a $\delta>0$ such that the following holds:
\begin{quote}
 Let $\mathscr{P}_1$, $\mathscr{P}_2$ be two partitions of $D$ together with choices of sampling points. If $\| \mathscr{P}_1 \| <\delta$ and $\| \mathscr{P}_2 \| <\delta$, then 
 $$
 \left| \mathcal{R}(f, \mathscr{P}_1) -  \mathcal{R}(f, \mathscr{P}_2) \right| <\epsilon.
 $$
 \end{quote}

(This property is used in the proof that $f$ is integrable on $D$.)

\medskip

Hint: let $\mathscr{P}$ be a common refinement of $\mathscr{P}_1$ and $\mathscr{P}_2$ -- that is, a partition that includes all the partition points of both $\mathscr{P}_1$ and $\mathscr{P}_2$. What can you say about
$\mathcal{R}(f, \mathscr{P}_1) -  \mathcal{R}(f, \mathscr{P})$ and $\mathcal{R}(f, \mathscr{P}_2) -  \mathcal{R}(f, \mathscr{P})$?}


\medskip

It suffices to prove the following. Let $\epsilon>0$. Then there is a $\delta>0$ such that if $\mathscr{P}_1$ is a partition of $D$ with $\| \mathscr{P}_1 \| <\delta$, and if $\mathscr{P}$ is a refinement of $\mathscr{P}_1$, then
\begin{equation}\label{hw3-e1}
 \left| \mathcal{R}(f, \mathscr{P}_1) -  \mathcal{R}(f, \mathscr{P}) \right| <\epsilon/2.
\end{equation}
Applying this to $\mathscr{P}_1$, $\mathscr{P}_2$ and their common refinement $\mathscr{P}$, we get
\begin{align*}
& \left| \mathcal{R}(f, \mathscr{P}_1) -  \mathcal{R}(f, \mathscr{P}_2) \right| 
\\
&\ \ \ \leq  \left| \mathcal{R}(f, \mathscr{P}_1) -  \mathcal{R}(f, \mathscr{P}) \right| 
+  \left| \mathcal{R}(f, \mathscr{P}) -  \mathcal{R}(f, \mathscr{P}_2) \right| 
\\
& \ \ \ <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
\end{align*}
We now prove (\ref{hw3-e1}). Let $D_{ij}=\{(x,y):\ x_{i-1}\leq x \leq x_i,\ y_{j-1}\leq y\leq y_j\}$ be a small rectangle in the partition $\mathscr{P}_1$, with the evaluation point $P_{ij}$. In the partition $\mathscr{P}$, this rectangle is subdivided further into smaller rectangles that we will call $R_{ijk\ell}$, with evaluation points $Q_{ijk\ell}$.

The contribution to $\mathcal{R}(f, \mathscr{P}_1) $ from $D_{ij}$ is
$$
f(P_{ij}) (\hbox{area of }D_{ij}) = \sum_{k,\ell} f(P_{ij}) (\hbox{area of }R_{ijk\ell}) .
$$
The contribution to $\mathcal{R}(f, \mathscr{P}) $ from $D_{ij}$ is
$$
\sum_{k,\ell} f(Q_{ijk\ell}) (\hbox{area of }R_{ijk\ell}) .
$$
The difference between the two contributions is
$$%\begin{equation}\label{hw3-e2}
\sum_{k,\ell} \Big(f(P_{ij}) - f(Q_{ijk\ell}) \Big) (\hbox{area of }R_{ijk\ell}).
$$%\end{equation}
But if $\mathscr{P}_1$ has diameter less than $\delta$, then the diagonal of each $D_{ij}$ is less than $\delta$. Since $P_{ij}$ and $Q_{ijk\ell}$ are two points in $D_{ij}$, their $x$-coordinates can differ by at most $\delta$, and similarly for their $y$-coordinates. Therefore 
$$
|f(P_{ij}) - f(Q_{ijk\ell})|\leq 2\delta
$$
so that
\begin{align*}
& \left| \sum_{k,\ell} \Big(f(P_{ij}) - f(Q_{ijk\ell}) \Big) (\hbox{area of }R_{ijk\ell})\right|
\\
& \ \ \ \leq 2\delta \sum_{k,\ell} (\hbox{area of }R_{ijk\ell})\\
&\ \ \ 
= 2\delta (\hbox{area of }D_{ij}).
\end{align*}
Adding up the $D_{ij}$ rectangles, we get
$$
\left| \mathcal{R}(f, \mathscr{P}_1) -  \mathcal{R}(f, \mathscr{P}) \right| 
 \leq 2\delta \sum_{i,j} (\hbox{area of }D_{ij}) = 2\delta.
$$
For this to be less that $\epsilon/2$, it suffices to take $\delta<\epsilon/4$.



\item {\it 
Let $D=\{(x,y):0\leq x\leq 1,0\leq y\leq 1\}$, and let $S=\{P_1,\dots,P_n\}\subset D$ be a set of finitely many points. Lef $f(x,y)$ be a function such that $f(P_1)=c_1$, ... , $f(P_n)=c_n$ for some arbitrary (but finite) values $c_1,\dots,c_n\in\rr$, and $f(x,y)=0$ for all $(x,y)\not\in S$.



Let $\mathcal{P}$ be a partition of $D$, with partition rectangles 
$D_{ij}=\{(x,y):x_{i-1}\leq x\leq x_i,y_{j-1}\leq y\leq y_j\}$.
Let $\mathcal{R}$ be the Riemann sum for $f$ associated with this partition.
 Let $d_{ij}=\sqrt{(x_i-x_{i-1})^2+(y_j-y_{j-1})^2}$ be the length of the diagonal of $D_{ij}$.  
 Assume that $0<d_{ij}<\epsilon$ for all $i,j$.
 Prove that 
$$
|\mathcal{R} |< C \epsilon^2
$$
for some constant $C$ that may depend on $n,c_1,\dots,c_n$, but is independent of $\epsilon$. What does this say about the integral $\iint_D f(x,y) \,dA$?}

\bigskip

Let $M=\max(|c_1|,|c_2|,\dots,|c_n|)$.

Since each rectangle of the partition
has diameter less than $\epsilon$, we have $0<x_i-x_{i-1}<\epsilon$ and $0<y_j-y_{j-1}<\epsilon$, so that each rectangle has area at most $(x_i-x_{i-1})(y_i-y_{j-1})<\epsilon^2$. 
Each point $P_k$ can belong to at most 4 rectangles of the partition (4 rectangles if $P_k$ is a ``corner point'', 2 if it lies on a common edge but is not a corner point, 1 otherwise). Therefore there are at most 4n rectangles that contain one or more points of $S$.  
The contribution to $\mathcal{R}$ from each such rectangle is at most $M \epsilon^2$, and the contricutions fron all other rectangles are zero. Therefore
$$
|\mathcal{R}|\leq 4n\cdot M\epsilon^2.
$$
Since this goes to zero as $\epsilon\to 0$, we have $\iint_D f(x,y) \,dA=0.$














\end{enumerate}

