My question is as follows.

If $(X,\mathcal A)$ and $(Y,\mathcal B)$ are two measure spaces and $\mathcal A_n\subseteq \mathcal A$ is a decreasing sequence of $\sigma$-algebras with $\mathcal A_n\downarrow \mathcal A_\infty$, then we have $$\mathcal A_n \times \mathcal B\downarrow \mathcal A_\infty \times \mathcal B$$

One direction is easy, but I am stuck at the other direction. Thank you for helping me with this.

I only did one direction as follows: take $A\times B\in \mathcal A_\infty \times \mathcal B$, we have that $A\times B\in \mathcal A_n\times \mathcal B$ for all $n$. So, $A\times B\in \bigcap_n(\mathcal A_n \times \mathcal B)$. The latter is a $\sigma$-algebra, so it contains $\mathcal A_\infty \times \mathcal B$, which is generated by the class of sets of the form $A\times B$. This implies $\mathcal A_\infty \times \mathcal B \subseteq \bigcap_n(\mathcal A_n \times \mathcal B)$.

We need to prove that $\bigcap_n(\mathcal A_n \times \mathcal B)\subseteq \mathcal A_\infty \times \mathcal B$.

Update: I should add another condition, that both $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ are probability spaces, and all the $\sigma-$ algebras are considered modulo the null sets. But I still want to know what happens if we don't have such a condition. Are there any counterexamples?

  • 3
    $\begingroup$ Why don't you tell us which direction you've already solved? $\endgroup$ – user940 Dec 18 '11 at 16:50
  • 3
    $\begingroup$ Are you sure this is actually true? (I'm not). $\endgroup$ – George Lowther Dec 18 '11 at 20:57
  • $\begingroup$ This is from the book "topics in ergodic theory", by William parry, page 71. I hope I didn't make any mistake copy it here. $\endgroup$ – Xiaochuan Dec 19 '11 at 17:50

It is not generally true that $\mathcal{A}_n\times\mathcal{B}$ decreases to $\mathcal{A}_\infty\times\mathcal{B}$. In general, the inequality $\mathcal{A}_\infty\times\mathcal{B}\subseteq\bigcap_n(\mathcal{A}_n\times\mathcal{B})$ holds, but the inequality can be strict. I'll demonstrate this by a counterexample.

Consider the case where $X=Y=2^\mathbb{N}$. An element $\omega$ of this space is a function $\omega\colon\mathbb{N}\to\{0,1\}$. Think of this as the space of an infinite sequence of coin tosses, where $\omega(n)=1$ if the n'th toss is a head and $\omega(n)=0$ if it is a tail. Also, let $\mathcal{A}=\mathcal{B}$ be the sigma-algebra generated by the individual tosses, so that it is generated by the sets $A_n=\{\omega\in X\colon\omega(n)=1\}$ ($n=1,2,\ldots$). Also, let $\mathcal{A}_n$ be the sigma-algebra generated by all the tosses on or after the n'th one. So, $\mathcal{A}_n$ is generated by $\{A_m\colon m\ge n\}$. Then, $\mathcal{A}_\infty=\bigcap_n\mathcal{A}_n$ is the tail sigma-algebra. I claim that the set $$ S=\bigcup_{n=1}^\infty\left\{(\omega_X,\omega_Y)\in X\times Y\colon \omega_X(m)=\omega_Y(m)\;{\rm for\ all\;}m\ge n\right\} $$ is in $\bigcap_n(\mathcal{A}_n\times\mathcal{B})$ but not in $\mathcal{A}_\infty\times\mathcal{B}$. Here, $S$ represents the pairs of sequences of coin tosses which agree with each other for all but finitely many tosses. The event for which they agree on every toss starting from the n'th one is $S_n=\bigcap_{m\ge n}((A_m\times A_m)\cup(A_m^c\times A_m^c))$, which is in $\mathcal{A}_n\times\mathcal{B}$. We then have $S=\bigcup_{m\ge n}S_m\in\mathcal{A}_n\times\mathcal{B}$ for all $n$, so $S\in\bigcap_n(\mathcal{A}_n\times\mathcal{B})$. In fact, $S\in\bigcap_n(\mathcal{A}_n\times\mathcal{A}_n)$.

We can also show that $S\not\in\mathcal{A}_\infty\times\mathcal{B}$. Let $\mathbb{P}$ be the probability measure on $(X\times Y,\mathcal{A}\times\mathcal{B})$ such that, the outcomes $(\omega_X,\omega_Y)\in X\times Y$ are distributed such that $\omega_X(n)$ are independent Bernoulli random variables with probability 1/2 of being equal to one (representing tossing a fair coin), $\omega_Y=\omega_X$ with probability 1/2, and $\omega_Y=1-\omega_X$ with probability 1/2. That is, if $\epsilon=(\epsilon_1,\epsilon_2,\ldots)$ is a sequence of independent random variables with $\mathbb{P}(\epsilon_k=1)=\mathbb{P}(\epsilon_k=0)=1/2$ then $$ \mathbb{P}(U)=\frac12\mathbb{P}\left((\epsilon,\epsilon)\in U\right)+\frac12\mathbb{P}\left((\epsilon,1-\epsilon)\in U\right) $$ for any $U\in\mathcal{A}\times\mathcal{B}$. Then, $\mathbb{P}(S)=1/2$. Now consider any set $U=V\times W\in\mathcal{A}_\infty\times\mathcal{B}$. Kolmogorov's zero-one law says that $V$ has probability 0 or 1. So, up to a set of zero probability, we have $U=X\times W$ or $U=\emptyset\times W=X\times\emptyset$. In any case, up to set of zero probability, $U$ can be written as $X\times W$ for some $W\in \mathcal{B}$ and, by the monotone class theorem (or pi-system d-system lemma), this extends to all $U\in\mathcal{A}_\infty\times\mathcal{B}$. However, $S$ is not of this form. In fact, as $S$ only depends on the events $C_n=\{(\omega_X,\omega_Y)\colon\omega_X(n)=\omega_Y(n)\}$, it can be seen that it is independent of $\{\emptyset,X\}\times\mathcal{B}$ so, if it did lie in this sigma-algebra (up to a probability zero event) then it would be independent of itself. This would imply $\mathbb{P}(S)=0$ or $\mathbb{P}(S)=1$, which is a contradiction.

It is true that, if $\mu,\nu$ are probability measures on $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ respectively and $\mathbb{P}=\mu\times\nu$ is the product measure, then every $S\in\bigcap_n(\mathcal{A}_n\times\mathcal{B})$ is in $\mathcal{A}_\infty\times\mathcal{B}$ up to a zero probability set. In fact, you can show that $$ \mathbb{E}\left[X\;\big\vert\mathcal{A}_n\times\mathcal{B}\right]\to\mathbb{E}\left[X\;\big\vert\mathcal{A}_\infty\times\mathcal{B}\right] $$ is probability as $n\to\infty$, for any integrable $\mathcal{A}\times\mathcal{B}$-measurable random variable $X$ (in fact, it converges almost-surely, but that is not needed here). In particular, if $X$ is $\bigcap_n(\mathcal{A}_n\times\mathcal{B})$-measurable, then the left hand side is just $X$, so $X=\mathbb{E}[X\;\vert\mathcal{A}_\infty\times\mathcal{B}]$ is (almost-surely) $\mathcal{A}_\infty\times\mathcal{B}$-measurable. To prove this limit, it is enough to consider $X=YZ$ where $Y$ $\mathcal{A}$-measurable and $Z$ is $\mathcal{B}$-measurable, and apply the monotone class lemma. You will probably need Levy's downward theorem to show that $\mathbb{E}[Y\;\vert\mathcal{A}_n]$ tends to $\mathbb{E}[Y\;\vert\mathcal{A}_\infty]$ in probability (I have a proof of the downward theorem on by blog, here).

  • $\begingroup$ I think the probability of $S$ above is also $0$. $\endgroup$ – Xiaochuan Dec 22 '11 at 23:24
  • $\begingroup$ @Xiaochuan: Why? By construction, the probability that $\omega_X(n)=\omega_Y(n)$ for all $n$ is 1/2. $\endgroup$ – George Lowther Dec 23 '11 at 22:55
  • $\begingroup$ Since the probability of $\omega_X(n)=\omega_Y(n)$ is $1/2$ for all $n$, then the probability of $\omega_X=\omega_Y$ is something like $\frac{1}{2}\frac{1}{2}\cdots=0$, not true? $\endgroup$ – Xiaochuan Dec 24 '11 at 11:00
  • $\begingroup$ @Xiaochuan. No, they're not independent. $\endgroup$ – George Lowther Dec 24 '11 at 11:45
  • $\begingroup$ oh yeah, I get you. $\endgroup$ – Xiaochuan Dec 24 '11 at 22:11

If $(X,{\mathcal A},P)$ and $(Y,{\mathcal B}, Q)$ are probability spaces, then $\cap_n({\mathcal A}_n\times{\mathcal B})$ is contained in the $P\times Q$-completion of ${\mathcal A}_\infty\times {\mathcal B}$. One way to prove this is by the martingale convergence theorem. (Exercise 6 on page 71 of Parry's book refers to the product of probability spaces, not just measurable spaces. Parry notes on page 8 of his text that inclusion between sigma-algebras is to be understood modulo null sets.)


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