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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)] $\ds f(x) = 1$ if $x$ is irrational,

$\ds f(x) = \sin \frac{1}{x}$ if $x$ is rational.
\item[(b)] $\ds g(x) = x$ if $x$ is irrational,

$\ds g(x) = \frac{b+1}{c}$ if $x$ is rational, and $\ds x =
\frac{b}{c}$ , where $b$ and $c$ have no common factors.
\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)] $$\begin{array}{lll} f(x) = 1 &amp; x{\rm \ irrational} &amp; 0&lt;x&lt;1 \\
f(x) = \sin \frac{1}{x} &amp;x{\rm \ irrational} &amp; 0&lt;x&lt;1\end{array}$$

$f$ is continuous if $\ds x = \frac{1}{(2n + \frac{1}{2})\pi} \,
\, n\epsilon{\bf N}$ and discontinuous elsewhere (proof is
standard analysis.)

Thus the set of discontinities of $f$ in (0,1) has measure 1 and
so $f$ is not R - integrable.  However $f=1$ a.e. in (0,1) and s
$f$ is Lebesgue integrable and $$L\int_0^1f =1$$

\item[(b)] $g$ is continuous at all irrational values (proof is
standard analysis)

So the set of discontinities has measure zero and so $g$ is R -
integrable.  Thus $g$ is L-integrable.  $g(x) = x$ a.e. and so
$$R\int_0^1g = L\int_0^1g = L\int_0^1x = \frac{1}{2}$$











\end{description}
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