Form a sequence $(En)_{n\ge0}$ of subsets of $R$ as follows: $E_{0} = [0, 1]$; $E_{1}$ is obtained from $E_{0}$ by removing the open middle interval of length $1/5$, i.e. $E_{1} = [0,2/5]∪[3/5, 1]$; $E_2$ is obtained by removing the open middle intervals of length $1/25$ from each of the $2$ intervals forming $E_{1}$, i.e. $E_{2} = [0,9/50 ] ∪ [11/50 ,2/5] ∪ [3/5,39/50 ] ∪ [41/50 , 1]$; In general, $E_{n}$ is obtained by removing the open middle intervals of length $1/5^{n}$ from each of the $ 2^{n−1}$ intervals forming $E_{n−1}$.
Let $E =\bigcap_{n=1}^{\infty} E_n$.
I'm trying to find the relation between $n$ and the length the of each $2^n$ interval ,
The only relation I could find is $L_{n}=(5^{n}-\sum_{k=1}^{n}5^{n-k}2^{k-1})/10^{n}$
Can any one help me with better relation , this one is not helping me to prove $E$ has zero measure .
They've pretty much already told you the main point: you start with measure $1$ and at the $n$th step you remove $2^{n-1}$ disjoint intervals each of which has measure $5^{-n}$. So the amount of measure that you remove is $\sum_{n=1}^\infty 2^{n-1} 5^{-n}$. This sum is easy to calculate using the geometric series.