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{\bf Question}

Discuss Riemann and Lebsgue integrability on (0,1) of the
functions $f$ and $g$ defined below.  In each case calculate the
integrals, where they exist.

\begin{description}
\item[(a)] $\ds f(x) = 0$ if $x$ is irrational,

$\ds f(x) = \frac{1}{c}$ if $x$ is rational and $\ds x =
\frac{b}{c}$ where $b$ and $c$ have no common factors.
\item[(b)] $\ds g(x) = 0$ if $x$ is rational,

$\ds g(x) = \frac{1}{a}$ if $x$ is irrational, where $a$ is the
first non -zero integer in the decimal representation of $x$.
\end{description}

(In each case you should prove any assertions you make concerning
continuity of function.  Conditions for integrability should be
stated but not proved.)

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]$$\begin{array}{ll} f(x) = 0 & x{\rm \ rational} \\ \ds
f(x) = \frac{1}{c} &\ds  x {\rm \ is\ rational\ and\ } x =
\frac{b}{c}
\end{array}$$
where $b$ and $c$ have no common factors.


We prove that $f$ is continious at each irrational point.  Let $x
\epsilon (0,1)$ be irrational.  Let $\epsilon>0$ be given. Choose
$\ds n > \frac{1}{\epsilon}.$


Since $x$ is irrational, there is an integer $m$ such that $$
\frac{m}{n!} < x< \frac{m+1}{n!} \hspace{.2in} 0 \leq m \leq n!$$
Let $\ds \delta = min\left\{ x - \frac{m}{n!}, \, \frac{m+1}{n!} -
x\right\}$

Consider the interval $I = (x - \delta , \, x + \delta)$

If $y\epsilon I $ and $y$ is irrational $|f(y) - f(x)| = 0
<\epsilon$

If $y\epsilon I $ and $y$ is rational then $\ds y = \frac{b}{c}, $
where $(b,c)$ = 1 and \underline{$c>n$}.  Hence $\ds |f(y) -
|f(x)| = \frac{1}{c} < \frac{1}{n} <\epsilon$.

Thus for all $y \epsilon (x-\delta, \, x+\delta) \, ,\ |f(y) -
f(x)| <\epsilon$ and so $f$ is continuous at $x$.

Thus $f$ is continuous almost everywhere and so is R - integrable.
Thus $f$ is also l-integrable and $$R\int_0^1f = L\int_0^1 f.$$
Now $f=0$ a.e. and so $$(L)\int_0^1f - (L)\int_0^1 o = 0$$ Hence
$$(R)\int_0^1 f = (L)\int_0^1f = 0$$

${}$

${}$


\item[(b)] $$\begin{array}{ll} g(x) = 0 & x{\rm \ irrational} \\
\ds g(x) = \frac{1}{a} & x {\rm \ is\ irrational\ and\ where\ } a
{\rm \ is the first\ non zero} \\ &{\rm didget\ in\ the\ decimal\
representation\ of\ }x \end{array}$$

We prove that $g$ is discontinuous at all irrational points.

Let $x \epsilon (0,1)$ be irrational, and $\ds g(x) = \frac{1}{a}
>0$.

Let $\ds \epsilon=\frac{1}{2a}$

Then for each $\delta >0$ there is a rational $y$ satisfying
$|y-x| = \delta$ and so $\ds |g(y) - g(x)| = \frac{1}{a}
>\epsilon$.

Hence $g$ is discontinuous at $x$.

Thus the set of discontinuations of $g$ in [0,1] form a set of
measure 1, and so $g$ is not Riemann integrable.

We now prove that $g$ is a simple function, and so is Lebesgue
integrable.

$g$ takes only the values $\ds 0, \frac{1}{9}, \frac{1}{8},
\frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{1}{3},
\frac{1}{2}, \frac{1}{1}$.

$\ds \{x| g(x) = 0\} = [0,1] \cap {\bf Q}$ and so has measure
zero, and is this measurable?  (${\bf Q}$ = set of rational
numbers)

$\ds \left\{x| g(x) = \frac{1}{a} \right\} = \bigcup_{n=1}^\infty
\left\{x| \frac{a}{10^n} \leq x \leq \frac{a+1}{10^n} \right\}
\cap C(Q)$

and this this measurable, being the union of a countable
collection of measurable sets intersected with a measurable set.

Also, since $m(Q) =0$, \begin{eqnarray*} m\left( \left\{ x| g(x) =
\frac{1}{a} \right\} \right) & = & m\left( \bigcup_{n=1}^\infty
\left\{x| \frac{a}{10^n} \leq x \leq \frac{a+1}{10^n} \right\}
\right) \\& =& \sum_{n=1}^\infty m\left(\left\{x| \frac{a}{10^n}
\leq x \leq \frac{a+1}{10^n} \right\} \right) \\ & = &
\sum_{n=1}^\infty \frac{1}{10^n} = \frac{1}{9} \end{eqnarray*}


Thus $g$ is a function of the form $$ g(x) = \sum_{i=1}^k c_i
X_{Ei}(x)$$ and so $$(L) \int_0^1 g = \sum c_i m(E_i) =
\frac{1}{9} \left( \frac{1}{9}+ \frac{1}{8}+ \frac{1}{7}+
\frac{1}{6}+ \frac{1}{5}+ \frac{1}{4}+ \frac{1}{3}+ \frac{1}{2}+
1\right)$$

$$ \left( = \frac{7129}{22680} = 0.314\right)$$








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