$P$ uniform distribution on $[0,1]$.
$$A_n=\bigcup_{i=1}^{2^n-1} \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ], n \in \mathbb{N}$$
To calculate $P(A_n)$ do we have to do the following??
$$P(A_n)=P \left ( \bigcup_{i=1}^{2^n-1} \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ] \right )=\sum_{i=1}^{2^n-1}P\left ( \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ] \right )=\sum_{i=1}^{2^n-1} \left ( \frac{2i}{2^n}-\frac{2i-1}{2^n} \right )=\sum_{i=1}^{2^n-1} \frac{1}{2^n} =\frac{1}{2^n} \sum_{i=1}^{2^n-1} 1=\frac{2^n-1}{2^n}$$