I have a question about an exercise problem from Durrett's probability book and its solution sketch in the solution manual.
On page 53, 2.1.18, it says
"If we want an infinite sequence of coin tossings, we do not have to use Kolmogorov's theorem. Let $\Omega$ be the unit interval $(0,1)$ equipped with the Borel set $\mathcal{F}$ and Lebesgue measure P. Let $Y_n(\omega) = 1 $ if $[2^n \omega]$ is odd and $0$ if $[ 2^n\omega]$ is even. Show that $Y_1, Y_2, ...$ are independent with $P(Y_k=0)=P(Y_k=1)=1/2.$"
In the solution manual, the solution is given in two lines as follows:
Let $i_1, i_2, ...\in \{0,1\}$ and $x = \sum_{m=1}^n i_m 2^{-m}$, then
$P(Y_1 = i_1,..., Y_n = i_n) =P(\omega \in [x,x+2^{-n})) = 2^{-n}. \quad*$
I understand that this solves the problem, but I cannot see how to justify the first equation in $*$. Could someone give a hint?
Note: To my knowledge, $[2^n \omega]$ is defined by $[x] = \max\{a \in \mathbb{Z}, a \leq x\}$ for $x\in \mathbb{R}.$
Ok, note that $x$ is a rational number with the specified binary expansion given by the $i_n$. What we want is all numbers who share those first $n$ digits in their binary expansion. Adding $2^{-n}$ will increment the $n$th digit in the expansion, but adding anything less than $2^{-n}$ will keep the first $n$ digits the same as $x$. Thus the interval we are interested in is $[x, x+2^{-n} )$, i.e. $\omega \in [x, x+2^{-n} )$ then the binary decimal expansion of $\omega$ has the first $n$ digits identical to those of $x$. The uniform probability of this is just the length of the interval.
Does that help? I can elaborate further if necessary.