Presently, I am reading Falconer's book on Geometric Measure theory. In example 4.2 he used mass distribution principle to calculate the lower bound of the Hausdorff dimension of Cantor set $C$. 
Now as he written $U$ can be intersect atmost one of the intervals in $E_k$. Now $U$ was an arbitrary open set whose measure is between $\frac{1}{3^{k+1}}$ and $\frac{1}{3^k}$ then it may so happen that $U$ intersect some part of one of intervals in $E_k$ and rest portion of $U$ lies outside. But then how one can claim
$ \mu(U)\leq 2^{-k}$ ??
Why we are only looking at the part of $U$ which is only intersecting $E_k$ whereas $U$ was any open set. Any clarification would be very helpful for my understanding.
The entire measure of the unit interval is carried by the Cantor set, $\mathscr{C}$. This means that if $E \subseteq [0,1]$, then $$\mu(E) = \mu( E \cap \mathscr{C}). $$ For any $k$, the Cantor set is entirely contained within $E_k$ and, by the mass distribution principle, each $E_k$ is made up of $2^{k}$ intervals (say $\{ I_{k,j} \}_{j=1}^{k})$, each of which has mass (measure) $2^{-k}$ and diameter (length) $3^k$. If $U$ is an open set with diameter between $3^{-(k+1)}$ and $3^{-k}$, then $U$ intersects at most one of the intervals making up $E_k$, say $$ U \cap E_k = U \cap I_{k,j} $$ for some $j$. Hence, by the monotonicity of the measure, $$ \mu(U) = \mu(U \cap \mathscr{C}) \le \mu(U \cap E_k) = \mu(U \cap I_{j,k}) \le \mu(I_{j,k}) = 2^{-k}. $$