# Infinite product of measurable spaces

1. Suppose there is a family (can be infinite) of measurable spaces. What are the usual ways to define a sigma algebra on their Cartesian product?
2. There is one way in the context of defining product measure on planetmath. Let $(E_i, B_i)$ be measurable spaces, where $i \in I$ is an index set, possibly infinite. We define their product as follows:

• let $E= \prod_{i \in I} E_i$ , the Cartesian product of $E_i$,

• let $B=\sigma((B_i)i \in I)$ , the smallest sigma algebra containing subsets of E of the form $\prod_{i \in I}B_i$ where $B_i=E_i$ for all but a finite number of $i \in I$ .

I was wondering why it is required that "$B_i=E_i$ for all but a finite number of $i \in I$"?

Thanks and regards!

I was wondering if the product sigma algebra defined in 2 is the smallest sigma algebra such that any tuple composed of one measurable set from each individual sigma algebra is measurable?

• Well, in 2 you give the usual way to define the product $\sigma$-algebra you ask for in 1... – Mariano Suárez-Álvarez Feb 14 '11 at 3:41
• Is 2 the only way? or the only useful way? – Tim Feb 14 '11 at 3:42
• A sensible $\sigma$-algebra on the product should be such that all projections $p_i:E\to E_i$ are measurable. This implies that any sensible $\sigma$-algebra contains the one you give in 2 and, therefore, the one you give in 2 is the smallest sensible $\sigma$-algebra. – Mariano Suárez-Álvarez Feb 14 '11 at 3:48
• Thanks! Can you explain why the projections $p_i$ being measurable requires that condition? – Tim Feb 14 '11 at 4:05
• @Tim: if you write down exactly what it means for the projections to be measurable, you'll see it yourself :) – Mariano Suárez-Álvarez Feb 14 '11 at 4:26

This is a good question. I haven't yet worked out a complete answer myself, but Mariano's comment above is definitely part of it: the given product $\sigma$-algebra has the property which is analogous to that of the product topology (which it resembles and is surely modelled on): it is the smallest $\sigma$-algebra which makes all of the projections measurable.

Because of this, I believe that if you make a category out of all measurable spaces and measurable functions in the obvious way, the product $\sigma$-algebra you have defined turns out to be the categorical product: i.e., it satisfies the requisite universal mapping property. Again, this is the situation for the product topology.

But I think the rest of the explanation has to do with the fact that this $\sigma$-algebra gives you the theorems you want, just as the product topology -- and not the "box topology" for instance -- has nice properties, especially Tychonoff's theorem. (The product topology was introduced for the first time in Tychonoff's paper, and his theorem played a large role in convincing mathematicians that it was the "right" topology on an infinite Cartesian product.)

I'm not sure exactly what the analogous result to Tychonoff's theorem is here, but I do know that this "coarsest" product $\sigma$-algebra enables one to define arbitrary products of probability spaces: see this lovely paper of S. Saeki for an incredibly short proof of that. I hope it is at least clear where the "coarseness" of the chosen product $\sigma$-algebra comes in handy: if (say in the case of a countably infinite index set, to fix ideas) (added: Michael Greinecker's answer shows that countable products actually behave rather well, so let's instead think about uncountable products) we allowed arbitrary products $Y = \prod_{i} Y_i$ of measurable subsets $Y_i \subset X_i$ to be measurable, then what should the measure of $Y$ be? If the set of indices for which $\mu(Y_i) < 1$ is uncountable, the product (taking the net of finite subsets of $I$) must approach zero. By requiring $Y_i = X_i$ for almost every $i$, we get that $\mu_i(Y_i) = 1$ for almost every $i$ and the infinite product is really a finite product.

Is there more to the story than this? Is the above construction of the product probability measure the "right" analogue of Tychonoff's theorem in this context (is there even a "right" analogue of Tychonoff's theorem in this context?)? I'm not sure, and I would be interested to hear more from others.

• I am not so sure about the universal property. Wasn't there a question on MO about how this fails? – Qiaochu Yuan Feb 14 '11 at 8:44
• @Qiaochu: I believe you are thinking of this question: mathoverflow.net/questions/49426. As it says (and as I said) the universal mapping property is in the category of measurable spaces, not the category of measure spaces. So I'm pretty sure what I said is true. – Pete L. Clark Feb 14 '11 at 9:05
• The issue is the following: The categorical product exists in the category of measurable spaces and the construction you give works. The product of probability spaces from measure theory is not a product in the category of measure spaces (whatever the morphisms you try). Dmitri Pavlov's answer to the question on MO gets around this by identifying two measure spaces if they have the same $L^{\infty}$-von Neumann algebra, and then expresses that in the (essentially) dual category of complete locales with distinguished $\sigma$-ideal. There are not many distinct objects left... – t.b. Feb 14 '11 at 13:40
• Essentially, the surprise boils down to the ancient observation (Hausdorff, von Neumann, Rokhlin) that every sufficiently nice probability space can be identified with part of the unit interval and at most countably many atoms. – t.b. Feb 14 '11 at 13:42
• @Pete: You would probably notice anyway, since you visit this site a lot. But since another answer was posted as an extensive comment to the answer by Pete L. Clark I thought it might be good to ping you. – Martin Sleziak Dec 23 '11 at 11:08

This is essentially an extensive comment on the post by Pete L. Clark.

We can define a product of a function with real values $(a_i)_{i\in I}$ in the following way (an elaboration on the definition by Pete): Let $\mathcal{F}$ be the set of finite subsets of $I$. It is a directed set under set inclusion. We define a net $\pi:\mathcal{F}\to\mathbb{R}$ by $\pi(F)=\times_{i\in F}a_i$. We let $\times_i a_i$ be the limit of the net $\pi$ if it exists. It is worth pointing out that this notion of convergence differs from the usual infinite product of a sequence of numbers in which convergence to zero is ruled out, so that convergence becomes equivalent to convergence of the series of logarithms of the factors. Essentially, the limit is then defined to be undefined when it would be zero otherwise. This makes little sense in probability theory, where we often want to show that a sequence of independent events has probability $0$. Also, the product defined here is "absolute", it does not depend on any order on the index set $I$.

Fact: If $(a_i)\in [0,1]^I$, then $\times_i a_i$ exists. If, moreover, $\times_i a_i>0$, then there exists a countable subset $C\subseteq I$ with $a_i=1$ for all $i\in I\backslash C$.

Proof: From the obvius fact that if $P\geq 0$ and $0\leq a_i\leq 1$ one has $Pa_i\leq P$, it follows that the net $\pi$ is nonincreasing. It is also obviously bounded below by $0$, so $\times_i a_i$ exists as the limit of a nonincreasing net that is bounded below. Assume now that $\times_i a_i>0$. For each positive natural number $n$, let $F_n=\{ i\in I:a_i<1-1/n \}$. For every $n$, $F_n$ is finite, so $\{i\in I:a_i<1\}=\bigcup_n F_n$ is countable.

Fact: If $I$ is countable and for every $i\in I$, $Y_i$ is a measurable subset of $X_i$, then $\prod_i Y_i$ is in the product $\sigma$-algebra.

Proof: W.l.o.g., let $I=\mathbb{N}$. For every $n$, the set $S_n=X_1\times\ldots X_{n-1}\times Y_n\times X_{n+1}\times\ldots=\pi_n^{-1}(Y_n)$ is by definition in the product $\sigma$-algebra. Now $\prod_{n\in\mathbb{N}}Y_n=\bigcap_{n=1}^\infty S_n$ which is in the $\sigma$-algebra as the countable intersection of measurable sets.

By a similar argument, one can show that for a general index set $I$, every measurable set can be written (subject to the usual labeling issues) as $\prod_{i\in C}Y_i\times\prod_{i\in I\backslash{C}}X_i$ with $C$ being a countable subset of $I$ and $Y_i$ an arbitrary measurable subset of $X_i$. So every measurable set in the product $\sigma$-algebra is determined by countably many coordinates. A useful Lemma for thinking about this and related issues is the following:

Lemma: Let $X$ be a set and $\mathcal{Y}$ be a family of subsets of $X$ and let $B\in\sigma(\mathcal{Y})$. Then there exists a countable family $\mathcal{C}\subseteq\mathcal{Y}$ such that $B\in\sigma(\mathcal{C})$.

Proof: Just verify that the family of all sets constructed by countable subfamilies forms a $\sigma$-algebra.

As a last remark: Tychonoff's theorem has a direct application in studying probabilities on infinite product spaces. The crucial step in proving the Daniell-Kolmogorov Extension Theorem is showing that the resulting probability measure is not just finitely additive but countably additive on the product algebra. For this one uses regularity with respect to the compact sets in the product topology (see here for a general approach). This is also the reason why the theorem does not hold for arbitrary product spaces (so the independent product of probability spaces is quite special).

• $@$Michael: if you can be a bit more explicit about the "two misleading mistakes" in my answer, I will try to address them. – Pete L. Clark Dec 24 '11 at 18:56
• $@$Michael: about the infinite product: not everyone agrees that an infinite product of positive numbers which approaches zero is "convergent". Restricting the sets as I did remedies this. – Pete L. Clark Dec 24 '11 at 18:59
• If you do probability theory, you have to allow convergence of a product to 0. How else are you going to calculate the probability of infinitely many independent coin-flips to be head in every case? – Michael Greinecker Dec 26 '11 at 23:45
• The other issue is that countable products of measurable sets are measurable. The countable product of (Y_n) is defined as the set of sequences $(y_n)$ with $y_n\in Y_n$ for all $n$. Each such condition defines a cylinder set and their intersection gives the product. In the paper by Saeki, he essentially works with the algebra generated to verify the conditions and makes the extension theorem do the work for the sigma-algebra. This does not mean that the latter does not contain countable products. – Michael Greinecker Dec 27 '11 at 0:00
• ...But it's clearly as a matter of taste. As for (2) -- i.e., that arbitrary countable products are automatically measurable because they are countable intersections of cylindrical sets...point taken! Thanks for making this explicit. – Pete L. Clark Dec 27 '11 at 5:30

Details of $$\textbf{lemma}$$ mencionated by Michael Greinecker : Let

$$L=\{E \subseteq X; E\in \sigma(C)\,\mbox{for some} \, C\subseteq Y,\,\mbox{with C countable}\}$$

See that the following hold:

$$\bullet$$ X, $$\emptyset \in L$$ (Obvious);

$$\bullet$$ If, $$E\in L$$, then $$E^{c}\in\sigma(C_E)$$ for $$C_E\subseteq Y$$ (countable) such that $$E\in\sigma(C_E)$$;

$$\bullet$$ if $$\{E_n\}_{n\in\mathbb{N}} \subseteq L$$, we can take $$\displaystyle C=\bigcup_{n=1}^{\infty}C_{E_n}$$ with $$C_{E_n}$$ in the same sense of above item. Then C is countable and $$\displaystyle\bigcup_{n=1}^{\infty}E_n \in \sigma(C)$$.

This proof that L is an $$\sigma$$-algebra. Finnaly observes that

$$\bullet$$ $$Y\subseteq L$$ (Using $$\sigma$$-algebra generating by each element of Y);

$$\bullet$$ For all countable $$C\subseteq Y$$ see that $$\sigma(C)\subseteq \sigma(Y)$$, and this implies that $$L\subseteq \sigma(Y)$$.

Because $$\sigma(Y)$$ is the smallest $$\sigma$$ -algebra containg Y, we have $$L=\sigma(Y)$$.

This proof that each event just depend of a countable many sets of generators.

$$\textbf{More about}:$$ In the case of product, by the observation of Michael you have that the product measure is generated by the sets $$\prod_{n\in C}Y_n \times \prod_{n\in I-C}X_n$$ with $$C\subseteq I$$ countable, this implies that for every $$E\subseteq X$$ exist an at most countable $$C\subseteq I$$ and $$E_C$$ in product sigma algebra of $$(X_i, B_i)_{i\in C}$$ satisfying $$E=\pi_{C}^{-1}(E_C)\hspace{3cm}(1)$$ for $$\pi_C : \prod_{i\in I} X_{i} \to \prod_{i\in C} X_{i}$$ the natural projection. This results says that, the product sigma algebra is not generated by the sets $$\prod_{i\in I} Y_{i}$$ with $$Y_i \in B_i$$ for all $$i\in I(uncontable)$$, indeed if every $$Y_i\neq X_i$$, for uncontable many of index $$i\in I$$, not exist an countable $$C\subset I$$ and $$E_C$$ such that (1) holds.