= Solution
{c}
For $1\leq j\leq n$, repeated use of the <floor function> identity gives
$$
X_j=2^{-j}\lfloor2^jX_n\rfloor.
$$
Thus the <filtration> satisfies $\mathcal F_n=\sigma(X_n)$. Each positive-<probability> <atom of a sigma-algebra> here has the form
$$
A_{n,k}=\{k2^{-n}\leq X<(k+1)2^{-n}\},\qquad 0\leq k<2^n,
$$
with <probability> $2^{-n}$; $\{X=1\}$ is a possible additional <null set> atom. Averaging over each <dyadic interval> gives a version
$$
\boxed{\mathbb E[h(X)\mid\mathcal F_n]
=2^n\int_{X_n}^{X_n+2^{-n}}h(u)\,du\quad\text{almost surely}.}
$$
For a globally defined version of the right side, assign any value, for example zero, on $\{X=1\}$.
Indeed, the displayed <random variable> is $\mathcal F_n$-<measurable> and bounded by $\sup|h|$. On each $A_{n,k}$ its <integral> is
$$
2^{-n}\,2^n\int_{k2^{-n}}^{(k+1)2^{-n}}h(u)\,du
=\mathbb E[h(X)\mathbf1_{A_{n,k}}],
$$
by the <uniform distribution>. Every member of $\mathcal F_n$ is a <union> of the displayed <sets> and the possible zero-<probability> atom. Summing these identities therefore verifies the defining <conditional expectation> property on all of $\mathcal F_n$.
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