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.