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Given a set $E$ which is bounded, measurable and $m^{\ast}(E)=x >0$, then for each $y \in (0,x)$ we can find a measurable set $A \subset E$, such that $m^{\ast}(A)=y$. To see this one considers measure as a continuous function and applies the Intermediate Value property.

Now we extend the question in the following manner: Suppose $E$ is a set which has finite measure, then for each positive $x < m^{\ast}(E)$ prove that there exists a perfect set $A \subset E$, such that $m^{\ast}(A)=x$.

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en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set - something like that? –  Asaf Karagila Oct 2 '10 at 13:40
Is $m^*$ a particular type of measure? The term is impossible to google, and I don't see it used in Kolmogorov & Fomin. –  aschepler Oct 2 '10 at 14:20
the general lebesgue measure –  anonymous Oct 2 '10 at 15:09
@Asaf: I have edited the question –  anonymous Oct 2 '10 at 15:11
By the continuity argument, I think he meant to look at $m(E\cap [0,y])$ as $y$ increases, which is continuous. –  JDH Oct 2 '10 at 17:36

1 Answer 1

up vote 3 down vote accepted

Suppose first that $E$ is contained in the closed interval $[a,b]$ and let $E' = [a,b] \setminus E$. By definition of Lebesgue measure, we can find an open set $G$ containing $E'$ such that $m(G) = m(E') + \varepsilon = (b - a) - m(E) + \varepsilon$, where $\varepsilon$ is small enough. Then $F = [a,b] \setminus G$ is a compact set contained in $E$ with measure $m(F) = m(E) - \varepsilon$.

Suppose we have chosen $\varepsilon$ small enough that $x \leq m(E)-\varepsilon = m(F)$. Your intermediate value argument proves the existence of some point $c \in [a,b]$ such that $m(F \cap [a,c]) = x$. By the Cantor-Bendixson Theorem, we have a decomposition $F \cap [a,c] = K \cup H$ where $K$ is perfect and $H$ is countable. Since $m(H) = 0$, it follows that $m(K) = m(F \cap [a,c]) = x$.

If $E$ is unbounded, then pick $[a,b]$ such that $m(E \cap [a,b]) = m(E) - \delta$, where $\delta$ is small enough, and apply the above argument to $E \cap [a,b]$.

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