Switching independent experiments: Does $I_A(X_1,\dots,X_n,Y_{n+1},Y_{n+2},\dots)$ converge almost surely?

Suppose there is a sequence of independent random variables $X=(X_1,X_2,\dots)$ with $X_i$ taking values in some arbitrary measure space $E_i$. Now there is a second sequence $Y=(Y_1,Y_2,\dots)$ independent of $X$ with the same distribution. Let $A\subseteq E_1\times E_2\times\cdots$ be any event. For $n\geq 1$ define $$I^A_n:=I((X_1,X_2,\dots,X_n,Y_{n+1},Y_{n+2},\dots)\in A):=\begin{cases}0~,~\text{if}~(X_1,\dots,X_n,Y_{n+1},Y_{n+2},\dots)\notin A\\ 1~,~\text{if}~(X_1,\dots,X_n,Y_{n+1},Y_{n+2},\dots)\in A,\end{cases}$$ so $I^A_n$ is an indicator. Now set $$I^A:=I((X_1,X_2,X_3,\dots)\in A):=\begin{cases}0~,~\text{if}~(X_1,X_2,X_3,\dots)\notin A\\ 1~,~\text{if}~(X_1,X_2,X_3,\dots)\in A\end{cases}.$$

For example consider the infinitely repeated toss of a fair coin. Imagine two people are tossing a coin independently with results $X_1,X_2,\dots$ and $Y_1,Y_2,\dots$. Here the event $A$ would be a Borel set of $\{0,1\}^{\mathbb{N}}$. For example $A=\{\text{Heads in the 27.th toss}\}.$ For each $n$ I built a new sequence of coin tosses: The first $n$ tosses come from $X$, the other infinitely many tosses come from $Y$. So the sequence of tosses is $(X_1,\dots,X_n,Y_{n+1},Y_{n+2},\dots)$. For the example $A=\{\text{Heads in the 27.th toss}\}$ you have $I^A_n=I(Y_{27}=1)$ for $n< 27$ and $I^A_n=I(X_{27}=1)=I^A$ for all $n>27$. Now in the general setting:

$$\text{Is it true that I^A_n converges almost surely towords I^A as n\rightarrow\infty?}$$ Since the RVs are $\{0,1\}$-valued convergence means that the RVs stay finally constant. I failed to prove this and I also failed to find a counterexample. I know the following:

1. $I^A_n$ and $I^A$ have the same distribution for all $n$.

2. The set $\mathcal{D}:=\{\text{Events$A$for which above statement is true}\}$ is closed under finite unions, intersections, taking complements and contains all cylindric events.

3. For every event $A$ the convergence $I^A_n\rightarrow I^A$ holds in $L^1$, so there is a deterministic subsequence $1\leq n_1<n_2<\dots$ depending on $A$ for which $I^A_{n_k}\rightarrow I^A$ hold almost surely as $k\rightarrow\infty$.

4. The event $\{I^A_n~\text{converges as$n\rightarrow\infty$}\}$ is contained in $\sigma(X)$ a.s.

5. The statement is true for any event $A$ with $P(X\in A)\in\{0,1\}$ so in particular for every terminal event, because of Kolmogorov-$0$-$1$-law.

6. Borel-Cantelli can't be used in general to prove almost sure convergence. There are examples where $\sum_{n\geq 1}P(|I^A-I^A_n|=1)=\infty$, but $I^A_n\rightarrow I^A$ almost surely holds anyway.

I failed to prove that $\mathcal{D}$ contains every event, so I tried to construct a counterexample: $E_i=\{0,1\}$ for every $i$ and $X$ is a sequence of independent fair coin tosses. By letting $U(x_1,x_2,\dots):=\sum_{i\geq 1}x_i/2^i$ and $C\subset[0,1]$ be a fat cantor set, i thought that $A:=\{(x_1,x_2,\dots):U(x_1,x_2,\dots)\in C\}$ gives a counterexample, but i failed to show that as well. See: https://mathoverflow.net/questions/242919/speed-of-convergence-in-lebesgues-density-theorem/242990#242990 there $2\cdot a_n(C)=P(|I^A-I^A_n|=1)$. The counterexample given there is not a counterexample for this question.

Do you know how to prove or how to give a counterexample?