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\begin{document}

\begin{center}
{\large SHEET 11: UNIFORM CONTINUITY AND INTEGRATION} \\ 
\end{center}

\bigskip \bigskip


%%---  sheet number for theorem counter
\setcounter{section}{11}   

We will now consider a notion of continuity that is stronger than ordinary continuity.  

\begin{definition}  Let $f \colon A \arr \bbR$ be a function.  We say that $f$ is \emph{uniformly continuous} if for all $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x, y \in A$
\[
\text{ if $\abs{x - y} < \delta$,} \quad \text{then $\abs{f(x) - f(y)} < \epsilon$.}
\]
\end{definition}

\begin{theorem}  If $f$ is uniformly continuous, then $f$ is continuous.
\end{theorem}

\begin{exercise}  
Determine with proof whether the following functions $f$ are uniformly continuous on the given
intervals $A$:
\begin{enumerate}
\item
$f(x)=x^2$ on $A=\bbR$

\item
$f(x)=x^2$ on $A=(-2,2)$

\item
$f(x)=\frac{1}{x}$ on $A=(0,+\infty)$

\item
$f(x)=\frac{1}{x}$ on $A=[1,+\infty)$

\item
$f(x)=\sqrt{x}$ on $A=[0,+\infty)$

\item
$f(x)=\sqrt{x}$ on $A=[1,+\infty)$
\end{enumerate}
%Find a function $f$ that is uniformly continuous.  Find a function that is continuous but not uniformly continuous.
\end{exercise}

\begin{exercise}
Let $p:\bbR\rightarrow\bbR$ be a polynomial with real coefficients.  Show that $p$ is uniformly
continuous on $\bbR$ if and only if $\mbox{deg}(p)\leq 1$.
\end{exercise}

\begin{exercise}
Let $f$ and $g$ be uniformly continuous on $A\subset \bbR$.  Show that:
\begin{enumerate}
\item  The function $f+g$ is uniformly continuous on $A$.

\item  For any constant $c\in \bbR$, the function $c\cdot f$ is uniformly continuous on $A$.

\end{enumerate}
\end{exercise}

%\noindent  
We will now prove that continuous functions with compact domain are automatically uniformly continuous.  To this end, first consider:

\begin{lemma}  
Let $f \colon A \arr \bbR$ be continuous.  Fix $\epsilon > 0$.  By the definition of continuity, for each $p \in A$ there exists $\delta(p) > 0$ such that for all $x \in A$
\[
\text{ if $\abs{x - p} < \delta(p)$,} \quad \text{then $\abs{f(x) - f(p)} < \tfrac{\epsilon}{2}$.}
\]
For each $p\in A$, define
$U(p) = \{ x \in \bbR \mid \abs{x - p} < \tfrac{1}{2} \delta(p) \}.$
Then the collection $\{ U(p) \mid p \in A \}$ is an open cover of $A$.
\end{lemma}



\begin{theorem}
Suppose that $X \subset \bbR$ is compact and $f \colon X \arr \bbR$ is continuous.  Then $f$ is uniformly continuous.
\end{theorem}

\begin{corollary}
Suppose that $f \colon [a, b] \arr \bbR$ is continuous.  Then $f$ is uniformly continuous.
\end{corollary}

\begin{definition}
We say that a function $f:A\to\bbR$ is \emph{bounded} if $f(A)$ is a bounded subset of $\bbR$.
\end{definition}
\begin{theorem}
Suppose that $X \subset \bbR$ is compact and $f \colon X \arr \bbR$ is continuous.  Then $f$ is bounded.
\end{theorem}

\begin{exercise}
Show that if $f$ and $g$ are bounded on $A$ and uniformly continuous on $A$, 
then $fg$ is uniformly continuous on $A$.
\end{exercise}

\medskip

We are now ready to turn to integration.  

\begin{definition} A \emph{partition} of the interval $[a, b]$ is a finite set of points in $[a, b]$ that includes $a$ and $b$:
\[
a = t_0 < t_1 < \dotsm < t_{n - 1} < t_n = b.
\]
If $P$ and $Q$ are partitions of the interval $[a,b]$ and $P\subset Q$, we refer to $Q$ 
as a \emph{refinement} of $P$.
\end{definition}

We usually write partitions as ordered lists $P = \{t_0, t_1, \dotsc, t_n\}$ with $t_{i - 1} < t_{i}$ for each $i=1,\ldots, n$.




\begin{definition}  Suppose that $f \colon [a, b] \arr \bbR$ is bounded and that $P = \{t_0, t_1, \dotsc, t_n\}$ is a partition of $[a, b]$.  Define:
\begin{align*}
m_i &= \inf \{ f(x) \mid t_{i - 1} < x < t_i \} \\
M_i &= \sup \{ f(x) \mid t_{i - 1} < x < t_i \}.
\end{align*}
The \emph{lower sum} of $f$ for the partition $P$ is the number:
\[
L(f, P) = \sum_{i = 1}^n m_i (t_i - t_{i - 1}).
\]
The \emph{upper sum} of $f$ for the partition $P$ is the number:
\[
U(f, P) = \sum_{i = 1}^n M_i (t_i - t_{i - 1}).
\]
\end{definition}

\noindent Notice that it is always the case that $L(f, P) \leq U(f, P)$.  

\begin{lemma}  Suppose that $P$ and $Q$ are partitions of $[a, b]$ and that $Q$ is a refinement of $P$.
%\footnote{in which case we say that $Q$ is a \emph{refinement} of $P$}.  
Then:
\[
L(f, P) \leq L(f, Q)  \quad \text{and} \quad U(f, P) \geq U(f, Q).
\]
\end{lemma}

\begin{theorem}  Let $P_1$ and $P_2$ be partitions of $[a, b]$ and suppose that $f \colon [a, b] \arr \bbR$ is bounded.  Then:
\[
L(f, P_1) \leq U(f, P_2).
\]
\end{theorem}

\begin{definition}  Let $f:[a,b]\rightarrow\bbR$.  We define:
\begin{align*}
L(f) &= \sup \{ L(f, P) \mid \text{$P$ is a partition of $[a, b]$} \} \\
U(f) &= \inf \{ U(f, P) \mid \text{$P$ is a partition of $[a, b]$} \}
\end{align*}
to be, respectively, the \emph{lower integral} and \emph{upper integral} of $f$ from $a$ to $b$.
\end{definition}

\begin{exercise}  Why do $L(f)$ and $U(f)$ exist?  Find a function $f$ for which $L(f) = U(f)$.  Find a function $f$ for which $L(f) \neq U(f)$.  Is there a relationship between $L(f)$ and $U(f)$ in general?
\end{exercise}

\begin{definition}  Let $f \colon [a, b] \arr \bbR$ be bounded.  We say that $f$ is \emph{integrable} on $[a, b]$ if $L(f) = U(f)$.  In this case, the common value $L(f) = U(f)$ is called the \emph{integral} of $f$ from $a$ to $b$ and we write it as:
\[
\int_{a}^{b} f.
\]
\end{definition}

When we want to display the variable of integration, we write the integral as follows, including
the symbol $dx$ to indicate that variable of integration:
\[
\int_{a}^{b} f(x) \, dx.
\]
For example, if $f(x)=x^2$, we would write 
$\displaystyle{\int_{a}^{b} x^2 \, dx}$ but not $\displaystyle{\int_{a}^{b} x^2}$.


\begin{theorem}\label{int_criterion}  Let $f \colon [a, b] \arr \bbR$ be bounded.  Then $f$ is integrable if and only if for every $\epsilon > 0$ there exists a partition $P$ of $[a, b]$ such that:
\[
U(f, P) - L(f, P) < \epsilon.
\]
\end{theorem}


\begin{theorem}\label{cont_implies_integrable}  If $f \colon [a, b] \arr \bbR$ is continuous, then $f$ is integrable.
\end{theorem}

\noindent (Hint: Use theorem \ref{int_criterion} and uniform continuity.)

\begin{exercise}  
Fix $c\in\bbR$ and let $f:[a,b]\rightarrow \bbR$ be defined by $f(x)=c$, for each $x\in [a,b]$.
Show that $f$ is integrable on $[a,b]$ and that $\int_a^bf=c(b-a)$.
\end{exercise}

\begin{exercise}  
Define $f:[0,b]\rightarrow \bbR$ by the formula $f(x)=x$.  
Show that $f$ is integrable on $[0,b]$ and that $\int_0^bf=\frac{b^2}{2}$.
\end{exercise}

\begin{exercise}  
Show that the converse of theorem \ref{cont_implies_integrable} is false in general.
\end{exercise}

\begin{exercise}
Let $f:[0,1]\rightarrow\bbR$ be defined by 
$$
f(x)=\left\{
\begin{array}{cl}
1 & \mbox{if $x\in \bbQ$} \\
0 & \mbox{if $x\not\in \bbQ$.} \\
\end{array}
\right.
$$
Show that $f$ is not integrable on $[0,1]$.  Compute the upper and lower integrals of $f$ on $[0,1]$.
\end{exercise}

\begin{theorem}  Let $a < b < c$.  A function $f \colon [a, c] \arr \bbR$ is integrable on $[a, c]$ if and only if $f$ is integrable on $[a, b]$ and $[b, c]$.  When $f$ is integrable on $[a,c]$, we have
\[
\int_{a}^{c} f = \int_{a}^{b} f + \int_{b}^{c}f.
\]
\end{theorem}

%\noindent 
If $b < a$, we define
\[
\int_{a}^{b} f = - \int_{b}^{a} f,
\]
whenever the latter integral exists.  With this notational convention, it follows that the equation
\[
\int_{a}^{c} f = \int_{a}^{b} f + \int_{b}^{c}f
\]
always holds, regardless of the ordering of $a$, $b$ and $c$, whenever $f$ is integrable on
the largest of the three intervals.









\begin{theorem}  Suppose that $f$ and $g$ are integrable functions on $[a, b]$ and that $c \in \bbR$ is a constant.  Then:
\begin{align*}
\hspace{-1.75in}&\mathrm{(i)} \quad  \int_{a}^{b} (f + g) = \int_{a}^{b} f + \int_{a}^{b} g, \\
\hspace{-1.75in}&\mathrm{(ii)} \quad \int_{a}^{b} c \cdot f = c \int_{a}^{b} f.
\end{align*}
\end{theorem}

\begin{theorem}  Suppose that $f$ is integrable on $[a, b]$ and that there exists numbers $m$ and $M$ such that:
\[
m \leq f(x) \leq M \qquad \text{for all $x \in [a, b]$.}
\]
Then:
\[
m (b - a) \leq \int_{a}^{b} f \leq M(b - a).
\]
\end{theorem}


\begin{theorem}  Suppose that $f$ is integrable on $[a, b]$.  Define  $F \colon [a, b] \arr \bbR$ by
\[
F(x) = \int_{a}^{x} f.
\]
Then $F$ is continuous.
\end{theorem}



\end{document}