In the probability space $((0,1),{\cal B}(0,1), \lambda)$, we define the Rademacher functions as:
$R_n(\omega)=\displaystyle\sum_{k=0}^{2^n-1}(-1)^{k+1}I_{\Delta_{k,n}}(\omega)$, where
$(0,1)=\displaystyle\bigcup_{k=0}^{2^n-1}\Delta_{k,n}$
$\Delta_{k,n}=\left(\dfrac{k}{2^n},\dfrac{k+1}{2^n}\right]$, for $k=0,1,\dots2^n-2$
and
$\Delta_{2^n-1,n}=\left(1-\dfrac{1}{2^n},1\right)$
I know that if $\omega\in(0,1)$ and $\omega=\displaystyle\sum_{n=1}^{\infty}\dfrac{x_n}{2^n}$ is the binary representation of $\omega$, then: $R_n(\omega)=-1\ \Longleftrightarrow\ $the $n^{th}$ binary digit of $\omega$ is 0 (i.e. $x_n=0$) $(I)$
and
$R_n(\omega)=1\ \Longleftrightarrow\ $the $n^{th}$ binary digit of $\omega$ is 1 (i.e. $x_n=1$) $(II)$.
Some prefer this as the definition of the Rademacher functions.
I would like to know why these two definitions are equivalent, that is why the equivalences $(I)$ and $(II)$ hold. Intuitively I understand it, but I can't seem to write it down as a rigorous mathematical proof.
Thanks in advance!
EDIT: I corrected the $\Delta_{k,n}$ sets. As it was, it wasn't a proper partition of $(0,1)$ (some elements were not included - the right endpoints of the $\Delta_{k,n}$'s).
The fact that $R_n(\omega)=1$ means that $\omega$ belongs to $\Delta_{k,n}$ for some odd integer $k$. That is, $k\leqslant2^n\omega\lt k+1$ and $k=2\ell+\color{red}{1}$ for some integer $\ell$. Hence $2^n\omega=2\ell+\color{red}{1}+s$ with $0\leqslant s\lt 1$. Since $0\leqslant k\lt2^n$, $0\leqslant\ell\lt2^{n-1}$ and $\ell$ is a binary integer $\ell=\sum\limits_{i=0}^{n-2}\ell_i2^i$ with $\ell_i=0$ or $\ell_i=1$. Thus, $$ \omega=\frac2{2^n}\sum\limits_{i=0}^{n-2}\ell_i2^i+\frac{\color{red}{1}}{2^n}+\frac{s}{2^n}=\sum\limits_{i=1}^{n-1}\frac{\ell_{n-i-1}}{2^i}+\frac{\color{red}{1}}{2^n}+\frac{s}{2^n}. $$ This is the binary expansion of $\omega$, up to and including its $n$th bit, and this $n$th bit is $\color{red}{1}$.
To solve the case $R_n(\omega)=-1$, rewrite the whole paragraph replacing each $\color{red}{1}$ by $\color{blue}{0}$.