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).


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