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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Let $T$ be the set of all real numbers in $(0,1)$ whose decimal
expansion does not contain the digit 7.  Prove that $T$ has
Lebesgue measure zero.

${}$

\item[b)]
Let $S$ be a measurable subset of $[0,1]$ with the property that
if $x\epsilon S$, and if the decimal expansion of $y$ differs from
that of $x$ in only a finite number of places then $y\epsilon S$.

${}$

Show that if $a$ and $b$ are real numbers in $[0,1]$ with
terminating decimal expansions, and $a<b$, then

$$m(S\cap[a,b])=m(S)\cdot(b-a).$$

${}$

Deduce that this result is true for all intervals
$[a,b]\subseteq[0,1]$.

${}$

Hence or otherwise prove that $S$ has measure 0 or 1.
\end{itemize}

\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Let $T$ be the set of all real numbers in $(0,1)$ whose decimal
expansion does not contain the digit 7.

Then $\ds T\subseteq\bigcup_{\begin{array}{c}r_i=0\\ r_i\not=7\\
i=1,\cdots
n\end{array}}^{10}\left[\sum_{i=1}^{n-1}\frac{r_i}{10^i}+\frac{r_n}{10^n},
\,\,\, \sum_{i=1}^{n-1}\frac{r_i}{10^i}+\frac{r_n+1}{10^n}\right]$

Thus $\ds m(T)\leq\sum_{\begin{array}{c}r_i=0\\ r_i\not=7\\
i=1,\cdots
n\end{array}}^{10}m\left[\sum_{i=1}^{n-1}\frac{r_i}{10^i}+\frac{r_n}{10^n},
\,\,\, \sum_{i=1}^{n-1}\frac{r_i}{10^i}+\frac{r_n+1}{10^n}\right]$

$\ds =\sum_{\begin{array}{c}r_i=0\\ r_i\not=7\\ i=1,\cdots
n\end{array}}^{10}\frac{1}{10^n}=\frac{9^n}{10^n}<\epsilon$ if $n$
is large enough.

Therefore $m(T)=0$.

${}$

${}$

\item[b)]
Suppose $\ds a=\frac{m_1}{10^{n_1}}, \,\,\,\,
b=\frac{m_2}{10^{n_2}}$

Let $n=\max(n_1,n_2)$ then we can write $\ds a=\frac{l_1}{10^n},
\,\,\,\, b=\frac{l_2}{10^n}$

$\ds [a,b]=\bigcup_{i=l_1}^{l_2-1}\left[\frac{i}{10^n}, \,\,\,
\frac{i+1}{10^n}\right]$

${}$

[These abutting intervals have just one point in common with their
neighbours, there are only a finite number of them, and so
additivity still holds.]

Now let $x\epsilon S\cap\left[\frac{i}{10^n}, \,\,\,
\frac{i+1}{10^n}\right]$ and consider $S\cap\left[\frac{j}{10^n},
\,\,\, \frac{j+1}{10^n}\right]$

The number $y=x+\frac{j-i}{10^n}$ will belong to
$\left[\frac{j}{10^n}, \,\,\, \frac{j+1}{10^n}\right]$ and will
differ at most the first $n$ decimal places from the expansion of
$x$.

Thus we shall have $y\epsilon S\cap\left[\frac{j}{10^n}, \,\,\,
\frac{j+1}{10^n}\right]$.

${}$

The reverse process can also be applied, and so this proves that

$S\cap\left[\frac{i}{10^n}, \,\,\, \frac{i+1}{10^n}\right]$ and
$S\cap\left[\frac{j}{10^n}, \,\,\, \frac{j+1}{10^n}\right]$

are simply translates of one another.

Thus $m(S)=10^n m\left(S\cap\left[\frac{i}{10^n}, \,\,\,
\frac{i+1}{10^n}\right]\right)$ for all $i$,

and so $\ds
m(S\cap[a,b[)=(l_2-l_1)m\left(S\cap\left[\frac{i}{10^n}, \,\,\,
\frac{i+1}{10^n}\right]\right)$

$=\frac{l_2-l_1}{10^n}m(S)=(b-a)m(S)$

${}$

Now if $a$ and $b$ are any real numbers $(a<b)$ we can express $a$
as a decreasing sequence of terminating decimals $\{a_n\}$ and $b$
as an increasing sequence of terminating decimals $\{b_n\}$ with
$b_n>a_n$.

${}$

Then $\ds (a,b)=\bigcup_{n=1}^\infty[a_n,b_n]$

and so $\ds S\cap(a,b)=\bigcup_{n=1}^\infty S\cap[a_n,b_n]$.

${}$

Thus $\ds m(S\cap[a,b])=m(S\cap(a,b))=m\left(\bigcup_{n=1}^\infty
S\cap[a_n,b_n]\right)$

$\ds
=\lim_{n\to\infty}m(S\cap[a_n,b_n])=\lim(b_n-a_n)m(S)=(b-a)m(S)$

${}$

Now let $I_n$ be any system of intervals satisfying $\bigcup
I_n\supseteq S$.

${}$

Then $m(S)=m(S\cap\bigcup I_n)=m(\bigcup S\cap I_n)$

$\leq\sum m(S\cap I_n)=m(S)\sum|I_n|$

i.e. $m(S)\leq m(S)\sum|I_n|$

so either $m(S)=0$ or $\sum|I_n|\geq1$

i.e. $m(S)\geq 1$ but $S\subseteq [0,1]$,

so either $m(S)=0$ or $m(S)=1$

\end{itemize}

\end{document}