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.
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.
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.
[0,1]=[0,a] U (a,1], now apply same idea for interval [0,a] like cantor's set construction.
