= Solution
For every $n$, $\mathcal F_n$ is a <sigma-algebra> contained in $\mathcal F$, and including one more <random variable> among its generators can only enlarge it. Hence $\mathcal F_n\subseteq\mathcal F_{n+1}$; the trivial $\mathcal F_0$ is contained in every <sigma-algebra>. These are exactly the requirements for a <filtration>.
The constant $X_0=x$ is $\mathcal F_0$-<measurable>. Suppose $X_n$ is $\mathcal F_n$-<measurable>. The pair $(X_n,U_{n+1})$ is then $\mathcal F_{n+1}$-<measurable>, and the <event> $\{U_{n+1}\leq X_n\}$ is $\mathcal F_{n+1}$-<measurable> because comparison of two real coordinates defines a <Borel set>. Define its <indicator function> by $B_{n+1}$. The recursion becomes
$$
X_{n+1}=\frac{X_n+B_{n+1}}2.
$$
It is a <measurable function> of that pair. <Mathematical induction> proves $X_n$ is $\mathcal F_n$-<measurable> for every $n$, so \b[$(X_n)$ is an adapted process]. The same recursion also proves $0<X_n<1$ at every finite time, a useful bound for the later <integrability> arguments.
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