The construction of the Cantor set begins with the removal of the interval $[1/3,2/3]$, an interval of length $1/3$. The remaining intervals have length $1/3$; you remove intervals of length $1/9$ from each of them. At stage $k$, you're removing $2^{k-1}$ intervals of length $1/3^k$. The total length removed is then $$\sum_{k=1}^\infty\frac{2^{k-1}}{3^k}=\frac{1}{3}\frac{1}{1-\frac{2}{3}}=1$$
so the Cantor set has measure zero.
Now let $\theta\ne 1/3$. You can repeat the same construction with $\theta$ instead of $1/3$: at step $k$, you remove $2^{k-1}$ intervals of length $\theta^k$. This is going to look very much like the Cantor set -- in particular, it has empty interior, which you can prove. If your homework problem is going where I'm guessing it's going, you should calculate the measure of the set you get for different $\theta$. Actually, you should do that no matter what the homework says. The results are quite surprising.