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{\bf Supplementary problems (assigned 11/3/03)}
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\vspace*{.3in}

\begin{enumerate}
\item 
Let $f:[a,b]\to\R$ be a bounded function which is continuous at all
points $x\in[a,b]$ except for those in a set $S\subset[a,b]$.  Suppose
that $S$ has the following property: \emph{for every $\epsilon > 0$,
there is a finite collection $I_1,\dots, I_n$ of disjoint
open intervals such that
\begin{itemize}
\item $S\subset I_1\cup\dots I_n$;
\item $|I_1| + \dots |I_n| < \epsilon$, where $|I_j|$ is the length of 
the interval $I_j$.
\end{itemize}}
Show that $f$ is Riemann integrable.  (Note that this implies that
a function which is continuous except on the Cantor set will be integrable!)
\vspace*{.2in}

\item 
Let $f:[0,1]\to \R$ be the restriction to $[0,1]$ of the function given
in problem 18 on page 100.  Show that $f$ is integrable.  

\emph{Remarks} (not part of the problem): nevertheless, $f$ does not satisfy 
the hypotheses of the previous problem---you'd need finitely many intervals 
that covered $\Q\cap[0,1]$, and it can be shown that the union of these 
intervals can omit only finitely many points in $[0,1]$.  Hence the sum of the
lengths of the intervals couldn't possibly be less than one.  
On the other hand, you can cover $\Q\cap[0,1]$ with \emph{countably} many
disjoint open intervals the sum of whose lengths is smaller than any given
$\epsilon$.  This need to allow for countably many intervals is one of the
keys to measure theory.
\vspace*{.2in}

\item (Integrals and series)
Suppose that we are given an infinite series $\sum_{k=1}^\infty a_k$.  
Suppose moreover that $a_k = f(k)$ where $f:[0,\infty)\to\R$ is a 
non-negative, decreasing function.   
\begin{itemize}
\item Show that $\int_{n-1}^m f(x)\,dx \geq \sum_{k=n}^m a_k
	\geq \int_n^{m+1} f(x)\,dx$.
\item Use this to show that $\sum_{n=2}^\infty \frac{1}{n(\log n)^2}$
converges.  (Note that the initial values of $n$
and $x$ are irrelevant---i.e. it's not a problem that our series starts 
with $n=2$ here.)
\item Note that the first item in this problem can be used to give an upper
bound on the difference between the full series and a given partial sum.
Use this bound to estimate how many terms of 
$\sum_{n=2}^\infty\frac{1}{n(\log n)^2}$ that you'd have to add up to be within
$.01$ of the sum of the full infinite series.  Assuming you had a computer
handy that was capable of adding up a trillion terms per second, how many
years would it take your computer to add up this many terms?
\item Now notice that you can use also use the first item to give a 
\emph{lower}
bound on the difference between the series and a given partial sum.  
These bounds allow you to get a good handle on the sum of the remaining
terms in the series.  Use this idea to estimate the value of the full 
series to within $.01$ in a more practical fashion (i.e. by Monday 11/10).
\end{itemize}

\end{enumerate}

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