Prove that $U$ is $\mathcal{A}-\mathcal{B}([0,1])$ measurable. Let's assume we're in the following context:

where cylinder set is $\left[\omega_1,...,\omega_n \right]=\{ \omega' \in \Omega: \omega'_i=\omega_i \}$.
I want to prove that $U$ is $\mathcal{A}-\mathcal{B}([0,1])$ measurable. I cannot use any convergence theorem like the Lebesgue's DCT.
I have no idea where to start. Maybe to prove that this function is continuous, but what distance would I use, and how would I prove it...
 A: Hints:


*

*Show that $X_n$ is $\mathcal{A}/\mathcal{B}([0,1])$-measurable for each $n \geq 1$.

*By step 1, $$U_k(\omega) := \sum_{n=1}^k X_n(\omega) 2^{-n}$$ is $\mathcal{A}/\mathcal{B}([0,1])$-measurable for all $k \geq 1$, and therefore $$U(\omega) = \lim_{k \to \infty} U_k(\omega)$$ is $\mathcal{A}/\mathcal{B}([0,1])$-measurable.

A: Lebesgue's DCT isn't applicable here since we're not looking at the convergence of integrals.
One way to show the measurability is to metrize the space of $\{0,1\}$ sequences with the following metric $d : \{0,1\}^{\mathbb{N}} \to [0,\infty)$,
\begin{align*}
d(a,b) = 2^{-n(a,b)}
\end{align*}
where $n(a,b)$ is the first integer $n$ at which the two strings $a$ and $b$ differ. Then you can show that the open sets induced by this metric generate the same $\sigma$ algebra as the one generated by the cylinder sets you mention.
From here it is easy to verify that the mapping you mention from binary expansions to real numbers in $[0,1]$ is continuous with respect to the metric given above (you should also verify that it is a metric). Thus the mapping is measurable.
A: Well, I'll just post an answer with the details from Daniel's answer.
we want $\displaystyle \forall_{y \in \Omega} \forall_{\epsilon>0} \exists_{\delta>0}\forall_{x \in \Omega} x\in B_{\delta}(y)\implies |U(x)-U(y)|<\epsilon$
For $x\neq y$, $x\in B_{\delta}(y)\Leftrightarrow d(x,y)=2^{-k}<\delta$, and 
$\displaystyle |U(x)-U(y)|<\sum^{\infty}_{i=k+1}2^{-i}=2^{-k}$, since first $k$ terms of the sum are equal. So if I choose $\delta=\epsilon$, then we have $ x\in B_{\delta}(y)\implies |U(x)-U(y)|<\delta=\epsilon$
