= Solution
Retain the binary <random variables> $B_j=\mathbf1_{\{U_j\leq X_{j-1}\}}$. Iterating the recursion gives
$$
2^nX_n=x+\sum_{j=1}^n2^{j-1}B_j.
$$
Each binary history determines the full vector $(X_0,\ldots,X_n)$, so its <sigma-algebra> is finite by the <sigma-algebra of a finite-valued random vector>. Conversely, because $0<x<1$, the integer $2^nX_n-x$ uniquely determines the digits $B_1,\ldots,B_n$; also $B_j=2X_j-X_{j-1}$. Thus
$$
\mathcal F_n^X=\sigma(B_1,\ldots,B_n)=\sigma(X_n).
$$
All $2^n$ binary histories have positive <probability>. By <independence> of the driving <uniform random variables>, along a fixed finite history every intermediate state is in $(0,1)$, so each subsequent conditional branch <probability>, either $X_{j-1}$ or $1-X_{j-1}$, is strictly positive. It follows that $\mathcal F_n^X$ has $2^n$ nonempty fibres, each an <atom of a sigma-algebra>, and exactly $2^{2^n}$ elements.
<Adaptedness> from the preceding part gives $\mathcal F_n^X\subseteq\mathcal F_n$. For $n\geq1$, however, $\mathcal F_n$ contains the <events> $\{U_1\leq t\}$ for every $t\in[0,1]$. They have distinct <probabilities> $t$, so there are infinitely many distinct such <events>. Consequently the finite <sigma-algebra> $\mathcal F_n^X$ cannot equal $\mathcal F_n$. \b[The inclusion is strict for every $n\geq1$.]
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