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{\bf Question}

Say what is meant by a Lebesgue measurable set in $R$ and show
that every set of Lebesgue outer measure zero is measurable.

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Define the Cantor ternary set, and show that it is uncountable.
Show also that it has measure zero.

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Give an example of a measurable function $f:[0,1]\rightarrow R$
and a Lebesgue measurable set $E\subseteq R$ for which $f^{-1}(E)$
is not a measurable set, proving your assertions.

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Hint: Use for $E$ a suitable subset of the Cantor set.

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[The existence of a non-measurable subset of $[0,1]$ may be
assumed.]



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{\bf Answer}

A set $E$ is measurable in $R$ iff for all $S\subset R, \,\,\,
m^*(S)=m^*(S\cap E)+m^*(S-E)$ where $m^*$ is Lebesgue outer
measure.  If $E$ has outer measure zero then since $S\cap
E\subseteq E, \,\,\, m^*(S\cap E)=0$.  Also $S-E\subseteq S$ and
so $m^*(S-E)\leq m^*(S)$.

Further, $S=(S\cap E)\cup(S-E)$

and so $m^*(S)\leq m^*(S\cap E)+m^*(S-E)=m^*(S-E)$

Hence $m^*(S)=m^*(S-E)=m^*(S-E)+m^*(S\cap E)$.

Thus $E$ is measurable.

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The Cantor ternary $K$ set is the set of numbers in $[0,1]$ which
have a decimal expansion in scale 3 containing only 0 or 2.
$m(K)=0$.

Suppose $x\epsilon K$ and $\ds x=\sum_{i=1}^\infty\frac{x_i}{3^i}$
where $x_i=0$ or 2.

Consider the mapping $f:K\rightarrow [0,1]$

$f:x\rightarrow y$ where $\ds
y=\sum_{i=1}^\infty\frac{\frac{1}{2}x_i}{2^i}$

The range of $f$ is $(0,1)$, since every $y\epsilon (0,1)$ has a
binary expansion.  Because of ambiguity over terminating and
non-terminating decimals $f$ is not $1-1$, but each
$y\epsilon(0,1)$ is the image of at most two $x\epsilon K$.  It
follows that $K$ is uncountable.

Now consider the function $g:[0,1]\rightarrow K$ defined as
follows.  If $x\epsilon[0,1]$ then $x$ has a unique
non-terminating binary expansion

$\ds x=\sum_{i=1}^\infty\frac{x_i}{2^i},\,\,\,$ define $\ds
g(x)=\sum_{i=1}^\infty\frac{2x_i}{3^i} \,\, \epsilon K$

$g$ is an increasing function and so is measurable, since
$\{x|f(x)<c\}$ is of the form $\{x|x<a\}$ or $\{x|x\leq a\}$.  Let
$F$ be a subset of $[0,1]$ which is not measurable.  Let
$E=g(F)\subseteq K, \,\,\,\, m^*(E)=0$ and so $E$ is measurable.
$g$ is $1-1$ and so $g^{-1}(E)=F$ which is not measurable.

This therefore provides the example required.




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