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Question: Show that if $0 \leq \text{m}(A) \leq \infty$, then for each positive $q < \text{m}(A)$ there is a perfect set $B \subset A$ of measure $q$. (All subsets are in $\mathbb{R}$ and the measure is the Lebesgue measure).

I can find a $B$ without the restriction of being perfect. Now for any given $\epsilon > 0 $ thre is a closed $F$ such that $\text{m}(B) - \text{m}(F) < \epsilon$. and so by Cantor Bendixson we get a perfect set with measure $q' = \text{m}(F)$ where $q - q' < \epsilon$ and this can be done for any $\epsilon$ but I am unable to find a perfect set with exact $q$ as the measure.

How to proceed?

This is problem $2.1.31$ from Kaczor and Nowak, Problems in Analysis III, Integration. I am unable to follow the solution given in the book.

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    $\begingroup$ Appropriately modify the construction process for any specific construction of a perfect nowhere dense set having positive measure. $\endgroup$ Oct 2 at 14:20
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    $\begingroup$ I am unable to follow up from your hint. I can make a nowhere dense perfect for any given measure $\alpha$ by translating the Cantor type sets. But how to ensure that the construction gives a subset inside $A$? I only know how to do it inside intervals. $\endgroup$ Oct 2 at 14:56
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    $\begingroup$ Sorry, I missed the part about doing it inside an arbitrary set of positive measure. What I said was intended for when $A$ is an interval of positive length. In case you're interested, see this MSE answer for similar results -- more general in some ways, less in other ways. $\endgroup$ Oct 2 at 14:59
  • $\begingroup$ @DaveL.Renfro Thank you for the link. The conjecture is interesting. for any $a < b$ we can make a set $A$ with $m_*(A)= a$ , $m^{*}(A) = b$(say using Vitali Sets), but with the added restriction of finding it in a given subset seems open. Of course in this $A$ cannot be perfect. $\endgroup$ Oct 4 at 13:04

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You can apply this countable procedure:

Since $m(A)>q$, for some $n_1$ sufficiently large $m(A)>q+\frac{1}{n_1}$. Now, choose $B\subseteq A$ arbitrary with $\mu(B)=q+\frac{1}{n_1}$ and set $\varepsilon_1:=\frac{1}{2n_1}$. We can find a closed set $F_{n_1}\subseteq B$ with measure at least $q+\frac{1}{2n_1}$ (and at most $q+\frac{1}{n_1}$), taking a countable set out we can assume that $F_{n_1}$ is perfect. Now take $F_{n_1}$ in place of $A$, the same procedure gives $F_{n_2}\subseteq F_{n_1}$ with measure between $q+\frac{1}{2n_2}$ to $q+\frac{1}{n_2}$, where $n_2>n_1$ (and we have complete freedom choosing $n_2$).

Continue this way (using induction) choosing $n_k\rightarrow\infty$, we get a sequence of perfect sets $...\subseteq F_{n_k}\subseteq F_{n_{k-1}}\subseteq...\subseteq F_{n_1}$.

The intersection of a decreasing sequence of perfect sets is a perfect set $F=\bigcap_{i=1}^\infty F_{n_i}$ and by construction and regularity of the Lebesgue measure, we have $m(F)=q.$

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  • $\begingroup$ Thank you, I also wanted to approximate from above to preserve the perfectness but this idea didn't strike me. $\endgroup$ Oct 3 at 1:47

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