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](https://mathoverflow.net/questions/243211/order-statistics-of-iid-uniform-rv-and-p%C3%B3lyas-urn-model-question-about-a-s-co)



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


