 Solution

For $m\ge 1$, fix signs $x_1,\dots ,x_m$. The constraints $X_j=x_j$ leave exactly two choices for the $(m+1)$ signs $Y_0,\dots ,Y_m$: choose $Y_0$ freely, then set $Y_1=Y_0{x}_1$ and $Y_j=x_{j-1}{x}_j$ for $2\le j\le m$. Each input tuple has probability $2^{-(m+1)}$ by independence. ConsequentlyEvery finite subfamily is a marginal distribution of one such initial segment, so the entire sequence is independent, with fair sign probability distributions.

Using the usual everywhere sign-valued representatives, for every $n\ge 1$ we haveThe 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) provesIn 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 $k\ge n$, $q_k$ 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.