Examples of perfect sets. Let $0\lt a\lt 1$. Can I get examples of of subsets of $[0,1]$ that are perfect sets, contains no intervals and has measure $1-a$.   
Well, I know by construction the Cantor set is perfect, contains no intervals. However, it has measure $0$. So it's no good.
 A: Unfortunately $\lim_{n \to \infty} (1-a/3)^n = 0$ so removing the middle $a/3$ does not work. As t.b.'s link shows, you have to very delicately remove smaller middle proportions the deeper you go, in order to retain perfection but with positive measure.
A: The construction at the link given by t.b. results in a Cantor set with measure $1/2$. At stage $n\ge 1$ it removes $1/4^n$ from the middle of each of the $2^{n-1}$ closed intervals left from the previous stage, so the measure of the deleted open intervals is $$\sum_{n=0}^\infty\frac{2^n}{4^{n+1}}=\sum_{n=0}^\infty\frac{2^n}{2^{2n+2}}=\sum_{n=0}^\infty\frac1{2^{n+2}}=\frac12\;.$$ If $\alpha\in(0,2)$, and the construction is modified to remove the middle $\alpha/4^n$ of each closed interval at stage $n$, the total amount removed will be $\alpha/2$, and the resulting Cantor set will have measure $1-\alpha/2$. Thus, it suffices to carry out the construction with $\alpha=2a$.
Alternatively, you could remove the middle $a/3^{n+1}$ from each closed interval at stage $n$, thereby removing a total of $$a\sum_{n=0}^\infty\frac{2^n}{3^{n+1}}=a\;;$$ this is what t.b. had in mind in his comment.
A: [0,1]=[0,a] U (a,1], now apply same idea for interval [0,a] like cantor's set construction. 
