Existence of Perfect subset for any given measure

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.

• Appropriately modify the construction process for any specific construction of a perfect nowhere dense set having positive measure. Oct 2 at 14:20
• 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. Oct 2 at 14:56
• 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. Oct 2 at 14:59
• @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. Oct 4 at 13:04

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 $$m(F)=q.$$