Calculate $P(A_n)$, where $P$ uniform distribution on $[0,1]$

Question

$P$ 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??

$$P(A_n)=P \left ( \bigcup_{i=1}^{2^n-1} \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ] \right )=\sum_{i=1}^{2^n-1}P\left ( \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ] \right )=\sum_{i=1}^{2^n-1} \left ( \frac{2i}{2^n}-\frac{2i-1}{2^n} \right )=\sum_{i=1}^{2^n-1} \frac{1}{2^n} =\frac{1}{2^n} \sum_{i=1}^{2^n-1} 1=\frac{2^n-1}{2^n}$$

Is this correct?? Or is it as followed??

$$P(A_n)=P \left ( \bigcup_{i=1}^{2^n-1} \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ] \right )=\sum_{i=1}^{2^n-1}P\left ( \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ] \right )=\sum_{i=1}^{2^n-1} \frac{1}{\frac{2i}{2^n}-\frac{2i-1}{2^n}}=\sum_{i=1}^{2^n-1} 2^n=2^n(2^n-1)$$

Answer 1

Your first calculation is correct. Note that

$$\sum_{i=1}^{2^n-1}P\left ( \left [ \frac{2i-1}{2^n}, \frac{2i}{2^n} \right ] \right )=\sum_{i=1}^{2^n-1} \frac{1}{\frac{2i}{2^n}-\frac{2i-1}{2^n}}$$

does not hold; that's where your second calculation fails.

Remark: You can simplifiy the calculation by noting that

$$A_n = \left[ \frac{1}{2^n}, 1 \right].$$