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\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $\sum_1^\infty \frac{1}{n^k}$ converges
        if and only if $k > 1$ ($*$). We saw that
        $\int_1^\infty \frac{1}{x^k}$ converges if
        and only if $k > 1$, so from the integral
        test we get ($*$).
    \item $\sum_2^\infty \frac{1}{n \log n}$,
        $f(x) = \frac{1}{x\log x}$, $x \ge 2$.
        \[ \int_2^R \frac{\dd x}{x \log x} =
        \left. \log(\log x) \right|_2^R =
        \log(\log R) - \log(\log 2) \to \infty \]
        as $R \to \infty$. Integral test implies
        \[ \sum_2^\infty \frac{1}{n \log n} \]
        diverges.
        \begin{corollary}[Euler's constant]
            As $n \to \infty$, $1 + \frac{1}{2} +
            \cdots + \frac{1}{n} - \log n \to
            \gamma$ with $0 \le \gamma \le 1$.
        \end{corollary}
        \begin{proof}
            Set $f(x) = \frac{1}{x}$ and use
            Theorem 5.14.
        \end{proof}

        \begin{note*}
            An open problem asks ``Is $\gamma$
            irrational? ($\gamma \approx 0.577$)''
        \end{note*}
\end{enumerate}

\myskip
We have seen: monotone functions and continuous
functions are Riemann integrable. We can
generalize this a bit and say that
\emph{piece-wise continuous} functions are
integrable.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/f106751ac4e311ec.png}
\end{center}

\begin{definition*}
    A function $f : [a, b] \to \RR$ is said to be
    piece-wise continuous if there is a dissection
    \[ \mathcal{D} = \{a = x_0, x_1, \dots, x_n =
    b\} \]
    such that
    \begin{enumerate}[(1)]
        \item $f$ is continuous on $(x_{j - 1},
            x_j) \,\,\forall\,\, j$
        \item The one-sided limits
            \[ \lim_{x \to x_{j - 1}^+} f(x),
            \qquad \lim_{x \to x_{j - 1}^-} f(x) \]
            exist.
    \end{enumerate}
\end{definition*}

\noindent
It is now an \emph{exercise} to check that $f$ is
Riemann integrable: just check that $f |_{[x_{j -
1}, x_j]}$ is integrable for each $j$. (the values
of $f$ and the endpoints won't really matter) and
use additivity of domain (property (6)).

\myskip
Question: How large can the discontinuity set of
$f$ be while $f$ is still Riemann integrable?

\myskip
Recall the example:
\[ f(x) = \begin{cases}
    \frac{1}{q} & x = \frac{p}{q} \\
    0 & \text{otherwise}
\end{cases} \]
on $[0, 1]$.

\begin{note*}
    What follows is non-examinable.
\end{note*}

\noindent
Answer: Henri Lebesgue characterization of Riemann
integrability: $f : [a, b] \to \RR$ bounded. Then
$f$ is Riemann integrable if and only if the set
of discontinuity points has \emph{measure zero}.

\begin{definition*}
    Let $l(I)$ be the length of an interval $I$.
    A subset $A \subset \RR$ is said to have
    measure zero if for each $\eps > 0$,
    $\exists$ a countable collection of intervals
    $I_j$ such that
    \[ A \subset \bigcup_{j = 1}^\infty I_j \]
    and
    \[ \sum_j l(I_j) < \eps \]
\end{definition*}

\begin{lemma*}
    \begin{enumerate}[(1)]
        \item Every countable set has measure
            zero.
        \item If $B$ has measure zero and $A
            \subset B$, then $A$ has measure zero.
        \item If $A_k$ has measure zero
            $\forall\,\, k \in \NN$, then
            $\bigcup_{k \in \NN} A_k$ also has
            measure zero.
    \end{enumerate}
\end{lemma*}

\subsubsection*{Oscillation of $f$}

$I$ interval:
\[ \omega_f(I) = \sup_I f - \inf_I f \]
oscillation of $f$ at a point:
\[ \omega_f(x) = \lim_{\eps \to 0} \omega_f(x -
\eps, x + \eps) \]

\begin{lemma*}
    $f$ is continuous at $x$ if and only if
    $\omega_f(x) = 0$.
\end{lemma*}

\begin{proof}
    Exercise.
\end{proof}

\subsubsection*{Brief Sketch of Lebesgue's criteria}

\[ D = \{x \in [a, b] : f \text{ discontinuous at
} x\} = \{x : \omega_f(x) > 0\} \]
\[ N(\alpha) = \{x : \omega_f(x) \ge \alpha\} \]
\[ D = \bigcup_1^\infty N \left( \frac{1}{k} \right) \]
Required to prove: $D$ has measure zero. Let $\eps
> 0$ be given, $\exists\,\, \mathcal{D}$ such that
\[ \sum_{j = 1}^n \omega_f([x_{j - 1}, x_j])(x_j -
x_{j - 1}) S(f, \mathcal{D}) - s(f, \mathcal{D}) <
\frac{\eps \alpha}{2} \]
\[ F = \{j : (x_{j - 1}, x_j) \cap N(\alpha) \neq
\emptyset\} \]
then for each $j \in F$,
\[ \omega_f([x_{j - 1}, x_j]) \ge \alpha \]
\[ \implies \alpha \sum_{j \in F} (x_j - x_{j -
1}) \le \sum_{j \in F} \omega_f([x_{j - 1},
x_j])(x_j - x_{j - 1}) < \frac{\eps \alpha}{2} \]
\[ \implies \sum_{j \in F} (x_j - x_{j - 1}) <
\frac{\eps}{2} \]
These cover $N(\alpha)$ except perhaps for $\{x_0,
x_1, \dots, x_n\}$. But these can be covered by
intervals of total length $< \frac{\eps}{2}$ hence
$N(\alpha)$ can be covered by total length $<
\eps$.

\myskip
For the other direction, let $\eps > 0$ be given.
$N(\eps) \subset D$, so $N(\eps)$ has measure
zero. $N(\eps)$ is closed and bounded hence it can
be covered by finitely many open intervals of
total length $< \eps$.
\[ N(\eps) = \bigcup_{i = 1}^m U_i \]
\[ K = [a, b] \setminus \bigcup_{i = 1}^m U_i \]
compact so it can be covered by finitely many
intervals $J_j$ such that
\[ \omega_f (J_j) < \eps .\]