Independent of $\sigma(X_{1})$ implies measurable with respect to $\sigma(X_{2},X_{3},\dots)$? Suppose $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space supporting iid random variables $\{X_{n}\}_{n \in \mathbb{N}}$.  Moreover, assume $\mathcal{F} = \sigma(X_{1},X_{2},\dots)$.  If $Y$ is a $\mathcal{F}$-measurable random variable and it is independent of $\sigma(X_{1})$, can we say that $Y$ is $\sigma(X_{2},X_{3},\dots)$-measurable?  
 A: Unless I messed up the counterexample below, the answer appears to be no.
Consider the case when $\mathbb{P}\{X_{1} = 0\} = \mathbb{P}\{X_{1} = 1\} = \frac{1}{2}$.  Define a random variable $Y$ by 
\begin{equation*}
Y(\omega) = h(X_{1}(\omega),X_{2}(\omega)),
\end{equation*}
where $h(x,y) = (1 - x) y + x (1 - y)$.  Observe that, if $j, k \in \{0,1\}$, then
\begin{align*}
\mathbb{P}\{Y = 0\} = \mathbb{P}\{Y = 1\} = \frac{1}{2} \\
\mathbb{P}\{Y = j, X_{1} = k\} = \frac{1}{4} = \mathbb{P}\{Y = j\} \mathbb{P}\{X_{1} = k\}
\end{align*}
Thus, $Y$ is independent of $X_{1}$.  
Nonetheless, $Y$ is not $\sigma(X_{2},X_{3},\dots)$-measurable since there is no measurable function $g : \{0,1\}^{\mathbb{N}} \to \mathbb{R}$ such that $Y = g(X_{2},X_{3},\dots)$ a.s.  
However, there is a version (i.e. a RV with the same distribution) of $Y$ on $(\Omega,\mathcal{F},\mathbb{P})$ that is $\sigma(X_{2},X_{3},\dots)$-measurable.  Indeed, $\tilde{Y}(\omega) = X_{2}(\omega)$ furnishes such a choice.
In fact, we have the following:

A $\sigma(X_{1},X_{2},\dots)$-measurable random variable $Y$ is independent of $\sigma(X_{1})$ only if there is a $\sigma(X_{2},X_{3},\dots)$-measurable random variable $\tilde{Y}$ such that $\tilde{Y} \overset{d}= Y$ (that is, $\tilde{Y}$ has the same distribution as $Y$).

To see this, observe that, since $Y$ is $\sigma(X_{1},X_{2},\dots)$-measurable, there is a Borel measurable function $h : \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$ such that $Y(\omega) = h(X_{1}(\omega),X_{2}(\omega),\dots)$.  I claim that the distribution of $Y$ conditioned on $X_{1} = x$ is almost surely independent of the choice of $x$.  
Indeed, if $A, B$ are Borel sets in $\mathbb{R}$ and $\mathbb{P}_{X_{1}}$ is the distribution of $X_{1}$, then, by independence,
\begin{equation*}
\mathbb{P}\{Y \in A\} \mathbb{P}\{X \in B\} = \mathbb{P}\left(\{Y \in A\} \cap \{X \in B\}\right) = \int_{B} \mathbb{P}\{Y \in A \, \mid \, X_{1} = x\} \mathbb{P}_{X_{1}}(dx).
\end{equation*}
Thus, there is a set $C \subseteq \mathbb{R}$ such that $\mathbb{P}_{X_{1}}(C) = 1$ and 
\begin{equation*}
\forall x \in C \quad \mathbb{P}\{Y \in A \, \mid \, X_{1} = x\} = \mathbb{P}\{Y \in A\}\end{equation*}
Therefore, the distribution of $Y$ conditioned on $X_{1} = x$ is constant as long as $x \in C$.
Next, we observe that $\mathbb{P}\{Y \in A \, \mid \, X_{1} = x\} = \mathbb{P}_{X_{2},X_{3},\dots}\{y \in \mathbb{R}^{\mathbb{N} \setminus \{1\}} \, \mid \, h(x,y) \in A\}$, where $\mathbb{P}_{X_{2},X_{3},\dots}$ is the joint distribution of $\{X_{2},X_{3},\dots\}$.  This follows from Fubini's Theorem.  Indeed, since we can write
\begin{equation*}
\mathbb{P}\{Y \in A\} = \int_{\mathbb{R}} \mathbb{P}_{X_{3},X_{3},\dots}\{y \in \mathbb{R}^{\mathbb{N} \setminus \{1\}} \, \mid \, h(x,y) \in A\} \, \mathbb{P}_{X_{1}}(dx),
\end{equation*}
it follows that $\mathbb{P}\{Y \in A \, \mid \, X_{1} =x\} = \mathbb{P}_{X_{2},X_{3},\dots}\{y \in \mathbb{R}^{\mathbb{N} \setminus \{1\}} \, \mid \, h(x,y) \in A\}$ for $\mathbb{P}_{X_{1}}$-almost every $x$.  
Finally, fix $x \in C$ and define $\tilde{Y}$ by $\tilde{Y}(\omega) = h(x,X_{2}(\omega),X_{3}(\omega),\dots)$.  Evidently $\tilde{Y}$ is $\sigma(X_{2},X_{3},\dots)$-measurable.  Moreover, it has the same distribution as $Y$.  Indeed, by Fubini's Theorem and the definition of $C$,
\begin{align*}
\mathbb{P}\{\tilde{Y} \in A\} &= \mathbb{P}\{h(x,X_{2},X_{3},\dots) \in A\} \\
&= \int_{\mathbb{R}} \mathbb{P}_{X_{2},X_{3},\dots}\{y \in \mathbb{R}^{\mathbb{N} \setminus \{1\}} \, \mid \, h(x,y) \in A\} \, \mathbb{P}_{X_{1}}(d\tilde{x}) \\
&= \int_{\mathbb{R}} \mathbb{P}\{Y \in A \, \mid \, X_{1} = x\} \mathbb{P}_{X_{1}}(d\tilde{x}) \\
&= \int_{\mathbb{R}} \mathbb{P}\{Y \in A \, \mid \, X_{1} = \tilde{x}\} \mathbb{P}_{X_{1}}(d\tilde{x}) \\
&= \mathbb{P}\{Y \in A\}.
\end{align*}
This concludes the proof of the "only if" direction. 
