= Solution
Write $\eta(A)=\int_A X_\infty\,d\lambda$. This is a finite positive <measure>. For $A\in\mathcal F_n$, the <conditional expectation> inequality gives
$$
\eta(A)=\int_A\mathbb E[X_\infty\mid\mathcal F_n],d\lambda\leq\int_A X_n\,d\lambda=\mu(A).
$$
The union $\mathcal A=\bigcup_n\mathcal F_n$ is an algebra of sets generating $\mathcal B([0,1))$, since half-open <dyadic intervals> generate all open sets and hence all Borel sets. The class $\{A:\eta(A)\leq\mu(A)\}$ is closed under increasing and decreasing limits, by continuity of these finite <measures>. The <Monotone class theorem> extends the inequality from $\mathcal A$ to every Borel set. Approximation by nonnegative simple functions and <monotone convergence> then give
$$
\boxed{\mu(f)\geq\int fX_\infty\,d\lambda\quad\text{for every nonnegative measurable }f.}
$$
The difference $\nu=\mu-\eta$ is therefore a finite nonnegative <measure>, not merely a signed <measure>. For $A\in\mathcal F_n$,
$$
\nu(A)=\int_A\left(X_n-\mathbb E[X_\infty\mid\mathcal F_n]\right)d\lambda.
$$
Consequently its restricted <Radon-Nikodym derivative> is
$$
\boxed{Y_n=X_n-\mathbb E[X_\infty\mid\mathcal F_n].}
$$
Choose the cell-average versions of all the <conditional expectations>. Then $Y_n$ is constant on each level-$n$ cell and equals $2^n\nu(D_k^n)$ there, so these versions are nonnegative at every point.
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