Dyadic expansion I'm reading the appendix in Billingsley book "Probability and measures" and I can't understand the following. If
$$\sum_{i=1}^n\frac{d_i(\omega)}{2^i} < \omega \leq \sum_{i=1}^n\frac{d_i(\omega)}{2^i} + \frac{1}{2^n},\qquad \forall n\geq 1,\forall \omega\in(0,1]$$
then
$\omega = \sum_{i=1}^\infty \frac{d_i(\omega)}{2^i}$
Here $d_1(\omega) = \begin{cases}0 & \omega\in(0,1/2]
 \\ 1 & \omega\in(1/2,1]\end{cases}$ and $d_i(\omega) = d_1(T^{i-1}(\omega))$ with $T(\omega) = \begin{cases}2\omega & \omega\in(0,1/2] \\ 2\omega - 1 & (1/2,1]\end{cases}$
In particular if this is true for any finite $n$, why we may pass to the limit and what happens with the first inequality, which is not strict?

$d_i(\omega)$ is the $i$th binary digit of $\omega$.
Let $S_n = \sum_{i = 1}^{n} \frac{d_i(\omega)}{2^i}$, and let $S = \sum_{i = 1}^{+\infty} \frac{d_i(\omega)}{2^i}$. 
Then 
$$S - S_n = \sum_{i = n+1}^{+\infty} \frac{d_i(\omega)}{2^i} \leq \sum_{i = n+1}^{+\infty} \frac{1}{2^i} \leq \frac{1}{2^n}.$$
It follows from this that
$$ S_n \leq S \leq S_n + \frac{1}{2^n}.$$
Since $\omega$ satisfies the same inequalities as $S$, we must have $|S - \omega| \leq \frac{1}{2^n}.$ As this is valid for all $n$, we must in fact have $\omega = S$. 
