3
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ I'm trying to understand the definition of $Z_n$. Specifically, what are the $X_n$'s? The initial part of your question suggests that $X_n$ is a function of $\omega$, so that $X_n(\omega)=\omega_n$. If this were the case, then of course $Z_n$ would be equal to $1_A(\omega)$ for all $n$. The interpretation that does make sense to me is if there is an $\omega$ and an $\omega'$. Is the question: is it true that for almost all $\omega$ and $\omega'$, is it true that $1_A(\omega_1,\ldots,\omega_n,\omega'_{n+1},\omega'_{n+2},\ldots)\to 1_A(\omega)$? Is this what you had in mind? $\endgroup$ – Anthony Quas Sep 1 '16 at 5:43
  • $\begingroup$ Yes, that's the question. Maybe I should have written $Z_n^{\omega}$ or $X_n(\nu):=\nu_n$ to avoid irritation. $\endgroup$ – user240643 Sep 1 '16 at 9:36
  • $\begingroup$ I added this formulation of the question. $\endgroup$ – user240643 Sep 1 '16 at 9:44
4
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.