= Solution
{c}
For $m\geq1$, fix signs $x_1,\ldots,x_m$. The constraints $X_j=x_j$ leave exactly two choices for the $(m+1)$ signs $Y_0,\ldots,Y_m$: choose $Y_0$ freely, then set $Y_1=Y_0x_1$ and $Y_j=x_{j-1}x_j$ for $2\leq j\leq m$. Each input tuple has <probability> $2^{-(m+1)}$ by <independence>. Consequently
$$
\mathbb P(X_1=x_1,\ldots,X_m=x_m)=2^{-m}
=\prod_{j=1}^m\mathbb P(X_j=x_j).
$$
Every finite subfamily is a <marginal distribution> of one such initial segment, so \b[the entire sequence is independent], with fair sign <probability distributions>.
Using the usual everywhere sign-valued representatives, for every $n\geq1$ we have
$$
Y_0=X_n\prod_{j=1}^nY_j.
$$
The right side is <measurable> with respect to $\sigma(\mathcal Y,\mathcal T_n)$. Therefore $Y_0$ is <measurable> with respect to their <intersection> $\mathcal G$.
Write $\mathcal T=\bigcap_n\mathcal T_n$. The <Kolmogorov zero-one law> applies to the <independent random variables> $(X_n)$, so every <event> $C\in\mathcal T$ has <probability> zero or one. Also $Y_0$ is <independent> of $\mathcal Y$: this follows first for finite <intersections> of sign <events> from the given <independence>, then for their generated <sigma-algebra> by part (a).
The <sets> $D\cap C$ with $D\in\mathcal Y$ and $C\in\mathcal T$ form a <pi-system> generating $\mathcal H$. If $\mathbb P(C)=0$, the <event> $D\cap C$ has <probability> zero and is <independent> of every <event> in $\sigma(Y_0)$. If $\mathbb P(C)=1$, it differs from $D$ by a <null set>, so the same <independence> follows from that of $Y_0$ and $\mathcal Y$. Another application of part (a) proves
$$
\boxed{Y_0\text{ is }\mathcal G\text{-measurable,}\qquad\sigma(Y_0)\text{ and }\mathcal H\text{ are independent}.}
$$
In particular $\mathcal H$ cannot contain the nontrivial <event> $\{Y_0=1\}$, whereas $\mathcal G$ does. Thus adjoining a <sigma-algebra> need not commute with a decreasing <intersection>.
There is a small version qualification if the assumptions specify only the <probability distributions>, allowing exceptional values other than signs. Put $q_n=X_n/(Y_1\cdots Y_n)$ when the denominator is nonzero, and $q_n=0$ otherwise. For $k\geq n$, $q_k$ is <measurable> with respect to $\sigma(\mathcal Y,\mathcal T_n)$. The <limit superior> of $(q_k)$, after <scalar clipping> to $[-1,1]$, is therefore $\mathcal G$-<measurable> and equals $Y_0$ <almost surely>. Literal <measurability> of an arbitrary version of $Y_0$ need not follow without choosing sign-valued representatives or adjoining the ambient <null sets>. All the <independence> conclusions are unaffected.
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