Given $$X_n\sim_\text{iid} \begin{bmatrix}0&1\\1/2&1/2\end{bmatrix}=\text{Bernoulli}(1,1/2)$$, and $S:=\sum_{n=1}^\infty \frac{X_n}{2^n}$, then, intuitively, $S$ has a uniform distribution on $[0,1]$, but how to formulate a rigorous proof?
One way found in Durrett is to note that the CF of the partial sum of $S$ converges to $\prod_{n=1}^\infty \cos(t/2^n)$, which is exactly $(\sin t)/t$, the CF of $U[0,1]$. However, this infinite trigonometric product is itself troublesome to deal with(turns out that it's quite easy..I don't understand why Durrett chooses to prove something easy using something hard, though) (indeed what Durrett does is the other way round, i.e. use the fact that $S\sim U[0,1]$ to prove this infinite product). So my question is, without invoking this infinite product, is there any way to prove $S$ is uniformly distributed in $[0,1]$?
$\newcommand{\Pr}{\mathbb{P}} $
I am assuming $\displaystyle S = \sum_{n=1}^{\infty} \frac{X_n}{2^n}.$
For any positive integer $n$ we have $2^n S = 2^n X_1 + 2^{n-1} X_2 + \dots + X_n + \displaystyle\sum_{k=1}^{\infty} \dfrac{X_{n+k}}{2^k} = I_n + F_n$ where $I_n = 2^n X_1 + 2^{n-1} X_2 + \dots + X_n$ and $F_n = \displaystyle\sum_{k=1}^{\infty} \dfrac{X_{n+k}}{2^k}.$
$I_n$ takes the values $0,1,\dots,2^n-1$ each with probablity $\dfrac{1}{2^n}$, and $\Pr(0 < F_n < 1) = 1$ ($F_n = 1$ would require all $X_k$'s to be $1$ for $k > n$ which has probability zero, $F_n=0$ is a zero probablity event for similar reasons).
Now choose any dyadic rational $r \in (0,1)$ of the form $r = \dfrac{k}{2^n}$,$1 \leq k \leq 2^n-1$ we have $\Pr(S \leq r) = \Pr( 2^nS \leq k) = \Pr(I_n = 0 ) + \dots + \Pr(I_n = k - 1) = \dfrac{k}{2^n}.$
(note $I_n = k$ implies $2^n S > k$ as $F_n > 0$ with probability 1)
So if $F$ is the distribution function of $S$ then $F(r) = r$ for any dyadic rational $r \in (0,1)$. Since the dyadic rationals are dense in $[0,1]$ the right continuity of $F$ ensures $F(x) = x$ for all $x \in [0,1].$