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{\bf Question}

Let $A$ be a closed measurable set of real numbers.  Prove that if
$A$ has an interior point then $A$ has positive measure.  Is the
converse true?  Justify your answer fully.

(Any properties of Lebesgue measure used should be stated
explicitly.)

\vspace{.25in}

{\bf Answer}


Suppose $A$ has an interior point $a$.

Then there exists $\epsilon >0$ such that $\ds (a-\epsilon,
a+\epsilon) \subseteq A$ $$m((a-\epsilon, a+\epsilon)) = 2\epsilon
>0,$$ and so, since if $\ds S \subseteq T$, $m(S) \leq m(T)$
$$m(A) \geq 2\epsilon >0$$ The converse if NOT true as the
following examples show.

Let $I_0$ be the unit interval.  Delete the middle open interval
of length $\lambda_1|I_0|$.  There remain tow closed interval,
denoted by $I_{1,1}$ and $I_{1,2}$ of length
$\frac{1}{2}(1-\lambda_1)$ each.

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\put(3,0.9){\line(0,1){.2}}

\put(.3,1.2){$I_{1,1}$}

\put(2.3,1.2){$I_{1,2}$}

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Let $\ds S_1 =\bigcup_{1=2}^2 I_{1,i}$

We delete from $\ds I_{11}$ and $\ds I_{12}$ the middle open
interval of length $\ds \lambda_2|I_{11}|$

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\put(6.2,0.9){\line(0,1){.2}}

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\put(1.5,1.2){$I_{2,2}$}

\put(4.3,1.2){$I_{2,3}$}

\put(5.5,1.2){$I_{2,4}$}

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There remain four closed intervals, denoted by $I_{2,1}$, $\ds
I_{2,2}$, $\ds I_{2,3}$ and $I_{2,4}$ each  of length $\ds
\frac{1}{2^2}(1 - \lambda_1)(1 - \lambda_2)$

Let $\ds S_2 = \bigcup_{i=1}^4 I_{2,i}$

We proceed inductively.

Let $\ds S_n = \bigcup_{i=1}^{2^n} I_{n,i}$ where $\ds |I_{n,i}| =
\frac{1}{2^n} \prod_{j=1}^n(1 - \lambda j)$, $i = 1, 2, \cdots
2^n; n = 1, 2, \cdots k$

We then remove the middle open interval of length $\ds
\lambda_{k+1}|I_{k,i}|$ from each $I_{k, i}$.  This leaves $\ds
2^{k+1}$ closed intervals $\ds I_{k+1, i}$, $ i = 1, 2, ...,
2^{k+1}$ each of length $\ds \frac{1}{2^{k+1}} \prod_{j=1}^{k+1}
(1 = \lambda_j)$.

Let $\ds s_{k+1} = \bigcup_{i=1}^{2^{k+1}} I_{k+1,i}$  Since, for
any closed interval, $m(I) = |I|$.and since $m$ is additive over
disjoint sets, we have \begin{eqnarray*} m(S_k) & = &
\sum_{i=1}^{2^k}m(I_{k,i}) \\ & = & \sum_{i=1}^{2^k}|I_{k,i}| \\ &
= & \sum_{i=1}^{2^k}\frac{1}{2^k} \prod_{j=1}^k ( 1 - \lambda_j)
\\ & = & \prod_{j=1}^k ( 1 - \lambda_j) \end{eqnarray*}

Also $\ds m(I_0) = 1 < \infty$, and $i_0\supseteq S_1\supseteq S_2
\supseteq \cdots$

Hence if we let $\ds S = \bigcap_{k=1}^\infty S_k$ $S$ is a
non-empty closed set and therefore measurable, and we have $\ds
m(s) = \lim_{k \to \infty}m(s_k) = \prod_{j=1}^\infty (1 -
\lambda_j)$ we choose $\ds \lambda_j = \frac{1}{2^j}$.  Then since
$\ds \sum \frac{1}{2^j}$ converges, $\ds \prod(1 - \frac{1}{2^j})$
converges to a number $\alpha$ with $\ds 0<\alpha<1$  Hence $\ds
m(S) = \alpha >0$.

It remains to prove that $S$ has no interior point.  Let $x$ be an
arbitrary point of $S$ and let $\epsilon>0$.  Choose $m$ so that
$\ds \frac{1}{2^{n-1}} < \epsilon$.  $\ds x \epsilon S$ so $\ds
x\epsilon S_n$. $\ds S_n = \bigcup_{i=1}^{2^n} I_{n,i}$ so for
some $i$ we have $\ds x\epsilon I_{n,i}$.

Let $y$ be the mid point of $I_{n-i,j}$, where $\ds j =
\frac{i}{2}$ or $\ds \frac{i+1}{2}$ according as $i$ is even or
odd.

$I_{n-i,j}$ is the interval giving rise to $\ds I_{n,i}$.  $y
\epsilon S_{n-1}$ but $\ds y \not\epsilon S_n$ and so $ y
\not\epsilon S$.

However $\ds x \epsilon I_{n-i,j}$ so $\ds | x - y| \leq
\frac{1}{2^{n-1}} < \epsilon$

Thus $x$ is not an interior point of $S$ and so $S$ as no interior
points.




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