Lebesgue Density Theorem: From convergence in probability to a.s. convergence

Question

Let $\Omega_1,\Omega_2,\dots$ be a sequence of finite nonempty sets endowed with discrete topology. The product space $$\Omega:=\Omega_1\times \Omega_2\times\cdots=\prod_{n\geq 1}\Omega_n$$ can be metrizised by the ultrametric $d((\omega_n),(\nu_n)):=\inf\{n:\omega_n\neq \nu_n\}^{-1}$ with $d((\omega_n),(\omega_n)):=0$. For $\varepsilon>0$ and $\omega\in\Omega$ let $$B_{\varepsilon}(\omega):=\{\nu\in\Omega:d(\omega,\nu)<\varepsilon\}.$$ So in particular for $\varepsilon=1/n$ you get $$B_{\varepsilon}(\omega)=\{\nu\in\Omega:\nu_1=\omega_1,\dots,\nu_n=\omega_n\}.$$

Now let $\mu_n$ be a probability measure on $\Omega_n$ and consider the product probability measure $\mu:=\otimes_{n\geq 1}\mu_n$ on $\Omega$. Let $X_n:\Omega\rightarrow \Omega_n,~\omega\mapsto \omega_n$ be the projection.

The paper (page $7$, proposition $2.10$) http://wwwmath.uni-muenster.de/u/ben.miller/papers/quasiinvariantmeasuresone.pdf states that Lebesgue density theorem holds (not only for product measures): For every Borelset $A\subseteq\Omega$ and $\mu$-a.e. $\omega\in\Omega$ $$\frac{\mu\left(B_{\varepsilon}(\omega)\cap A\right)}{\mu\left(B_{\varepsilon}(\omega)\right)}~\longrightarrow~1_A(\omega)~~\text{as}~~\varepsilon\rightarrow 0,$$ where $1_A(\cdot)$ denotes the indicator function of the set $A$.

Now for $\varepsilon=1/n$ you get because of the independence of $X_1,X_2,\dots$ $$\frac{\mu\left(B_{\varepsilon}(\omega)\cap A\right)}{\mu\left(B_{\varepsilon}(\omega)\right)}~=~\mu\big((\omega_1,\dots,\omega_n,X_{n+1},X_{n+2},X_{n+3},\dots)\in A\big)~=~\mu\bigg(1_A((\omega_1,\dots,\omega_n,X_{n+1},X_{n+2},\dots))=1\bigg).$$ For $\mu$-a.e $\omega\in\Omega$ and every $\delta>0$ you get $$\mu\big(|1_A((\omega_1,\dots,\omega_n,X_{n+1},X_{n+2},\dots))-1_A(\omega)|>\delta)\longrightarrow 0~~\text{as}~~n\rightarrow\infty.$$

So the Lebesgue-Density-Theorem implies that for $\mu$.a.e $\omega\in\Omega$ the $\{0,1\}$-valued sequence $(Z_n)$ with $$Z_n:=1_A((\omega_1,\dots,\omega_n,X_{n+1},X_{n+2},\dots))$$ converges in probability w.r.t $\mu$ towards the constant $1_A(\omega)$ as $n\rightarrow\infty$.

Since almost sure convergence implies convergence in probability, here is the

Question:

Let $A\subseteq\Omega$ be a Borelset. Is it true that for $\mu$-a.e. $\omega\in\Omega$ the sequence $(Z_n)$ converges $\mu$-almost-surely towards $1_A(\omega)$ as $n\rightarrow\infty$?

Or in other terms: Is it true that for $\mu\otimes\mu$-a.e. $(\omega,\omega')\in\Omega\times\Omega$
$1_A(\omega_1,\dots,\omega_n,\omega_{n+1}',\omega_{n+2}',\dots)$ converges towards $1_A(\omega)$ as $n\rightarrow\infty$?

Since the sequence under consideration is $\{0,1\}$-valued a.s convergence means that the sequence stays finally constant $\mu$-a.s.

Remark

A positive answer would imply a positive answer to this unanswered question of mine, which is a special instance of the above situation: Order statistics of iid uniform RV and Pólya's urn model. Question about a.s. convergence

Answer 1

The answer is "No, of course". Let us denote $(\omega,\omega',n)=(\omega_1,\dots,\omega_n,\omega'_{n+1},\dots)$. Take any large $m$ and consider the event $A_m$ that $\sum_{n=0}^{2m}\omega_n=m$. Then $P(A_m)\approx m^{-1/2}$ but for every $n_0$, we have $P((\omega,\omega',n)\in A_m\text{ for some }n\ge n_0)\ge\frac 13$ if $m$ is large enough (it is enough that $\sum_{n=0}^{2m}\omega'_n$ deviate from $m$ by at least $n_0$ and the same sum for $\omega$ be on the other side of $m$). Now just take some sequence $m_j$ so that $\sum_j m_j^{-1/2}$ is very small and take $A=\cup_j A_{m_j}$. Then $P(A)$ is almost $0$ while the probability that there are infinitely many $1$'s in the sequence $1_A((\omega,\omega',n))$ is $1$ (it is obvious that it is $\ge \frac 13$, but it is a tail event).