Events That Inspect Only Finitely Many Coordinates
In a countable product, many events of interest ask about only a finite number of coordinates: for example, whether coordinate 2 lies in one set while coordinate 7 lies in another. Such events can be described before a probability measure on the whole infinite product has been constructed. The key is to encode each event in a finite-dimensional space, and use consistency to ensure that its probability does not depend on which finite-dimensional representation is chosen.
Let \((S_i,\mathcal{A}_i)\), \(i\in\mathbb{N}\), be nonempty measurable spaces, and let \(X=\prod_{i=1}^{\infty}S_i\). For a finite index set \(F\subset\mathbb{N}\), write \(S_F=\prod_{i\in F}S_i\), with sigma-algebra \(\mathcal{A}_F=\bigotimes_{i\in F}\mathcal{A}_i\), and let \(\pi_F:X\to S_F\) be the coordinate projection. For finite \(F\subset G\), write \(\pi_{G,F}:S_G\to S_F\) for the projection from the larger finite product to the smaller one. We take \(S_{\varnothing}\) to be a one-point space.
Each cylinder belongs to the product sigma-algebra \(\bigotimes_{i=1}^{\infty}\mathcal{A}_i\), since the coordinate projections are measurable. The cylinder algebra is an algebra of subsets: it contains \(X\) and is closed under complements and finite unions. An algebra need not be a sigma-algebra; closure under countable unions is not part of its definition.
Why Cylinders Form an Algebra
Proof. The whole space is the cylinder \(C(\varnothing,S_{\varnothing})\). Consider a cylinder \(C(F,A)\). Its complement is \(X\setminus\pi_F^{-1}(A)=\pi_F^{-1}(S_F\setminus A)\), and \(S_F\setminus A\in\mathcal{A}_F\), so the complement is a cylinder on \(F\).
Now take finitely many cylinders \(C(F_j,A_j)\), \(j=1,\ldots,r\), and put \(H=\bigcup_{j=1}^{r}F_j\). This is a finite set. For each \(j\), the set \(D_j=\pi_{H,F_j}^{-1}(A_j)\) belongs to \(\mathcal{A}_H\), because the coordinate projection \(\pi_{H,F_j}\) is measurable. The projections satisfy \(\pi_{F_j}=\pi_{H,F_j}\circ\pi_H\), so \(C(F_j,A_j)=\pi_H^{-1}(D_j)\). Consequently, \[ \bigcup_{j=1}^{r}C(F_j,A_j)=\pi_H^{-1}\left(\bigcup_{j=1}^{r}D_j\right), \] and the set inside the parentheses is in \(\mathcal{A}_H\). Thus the union is a cylinder. The same common-coordinate representation also gives intersections by replacing the union of the \(D_j\)'s with their intersection. This proves the proposition. \(\square\)
Worked Example: Combining Two Coordinate Conditions
Let \(S_i=\{0,1\}\) with all subsets measurable. Consider \(C_1=\{x\in X:x_1=1\}\) and \(C_2=\{x\in X:x_3=0\}\). Express their union and intersection using the common coordinate set \(H=\{1,3\}\).
In \(S_H=\{0,1\}^2\), with coordinates written in the order \((x_1,x_3)\), the corresponding finite-dimensional events are \(D_1=\{(1,0),(1,1)\}\) and \(D_2=\{(0,0),(1,0)\}\). Therefore \(C_1\cup C_2=\pi_H^{-1}(\{(1,0),(1,1),(0,0)\})\), while \(C_1\cap C_2=\pi_H^{-1}(\{(1,0)\})\). For the complement, \(C_1^c=\pi_{\{1\}}^{-1}(\{0\})\). These descriptions impose no condition on coordinates outside the displayed finite sets.
Check: among the four points in \(\{0,1\}^2\), the union event contains three points, and the intersection contains only \((1,0)\). The descriptions agree directly with the conditions “\(x_1=1\) or \(x_3=0\)” and “\(x_1=1\) and \(x_3=0\).”
The cylinder algebra is generally smaller than the product sigma-algebra. For instance, take \(S_i=\{0,1\}\) with all subsets measurable and let \(C_n=\{x:x_n=1\}\). The union \(\bigcup_{n=1}^{\infty}C_n\) consists of sequences having at least one coordinate equal to 1. It is not a cylinder: if it depended on a finite set \(F\), choose \(k\notin F\). The all-zero sequence and the sequence with value 1 only at coordinate \(k\) agree on \(F\), but only the second belongs to the union. Thus the cylinder algebra is not closed under countable unions in this example.
Assigning Probabilities from a Consistent Family
Suppose that for every finite \(F\subset\mathbb{N}\) we are given a probability measure \(\mu_F\) on \((S_F,\mathcal{A}_F)\). These finite-dimensional laws are consistent if, whenever \(F\subset G\) are finite and \(A\in\mathcal{A}_F\), \(\mu_G(\pi_{G,F}^{-1}(A))=\mu_F(A)\). The law on \(S_{\varnothing}\) is the probability measure assigning mass 1 to its single point. Consistency says that taking a marginal of a larger finite-dimensional law gives the specified law on the smaller set of coordinates.
Proof. First, check that the formula does not depend on the representation of a cylinder. Suppose \(C(F,A)=C(G,B)\), and let \(H=F\cup G\). Define \(A_H=\pi_{H,F}^{-1}(A)\) and \(B_H=\pi_{H,G}^{-1}(B)\); these are measurable subsets of \(S_H\). Their inverse images under \(\pi_H\) are the same cylinder. The map \(\pi_H:X\to S_H\) is surjective: any point of \(S_H\) can be extended to a point of \(X\) by choosing an element in each nonempty \(S_i\) for \(i\notin H\). Hence \(A_H=B_H\). By consistency, \[ \mu_F(A)=\mu_H(A_H)=\mu_H(B_H)=\mu_G(B). \] The assigned value is therefore well-defined.
The whole space has probability \(\mu_{\varnothing}(S_{\varnothing})=1\), and the empty cylinder has probability 0. To prove finite additivity, let \(C_1,\ldots,C_r\) be pairwise disjoint cylinders, represented on a common finite set \(H\) as \(C_j=\pi_H^{-1}(D_j)\). The sets \(D_j\) are pairwise disjoint: if some point belonged to \(D_j\cap D_k\), surjectivity of \(\pi_H\) would give a point in \(C_j\cap C_k\). Finite additivity of the probability measure \(\mu_H\) now gives \[ \mu_0\left(\bigcup_{j=1}^{r}C_j\right) =\mu_H\left(\bigcup_{j=1}^{r}D_j\right) =\sum_{j=1}^{r}\mu_H(D_j) =\sum_{j=1}^{r}\mu_0(C_j). \] This proves normalization and finite additivity on the cylinder algebra. \(\square\)
Worked Example: A Consistent Bernoulli Family
For each finite \(F\subset\mathbb{N}\), define a probability law on \(\{0,1\}^F\) by \(\mu_F(\{z\})=\prod_{i\in F}(1/3)^{z_i}(2/3)^{1-z_i}\) for each point \(z=(z_i)_{i\in F}\), and extend it to all subsets by summing point masses. Verify consistency and compute the probability that \(x_2=1\) and \(x_5=0\).
For a fixed assignment on \(F\), summing over the value \(z_j\) of any additional coordinate \(j\notin F\) multiplies its mass by \((2/3)^{1}+(1/3)^{1}=1\), where the two terms are the probabilities for \(z_j=0\) and \(z_j=1\), respectively. Thus marginalizing any added coordinates recovers \(\mu_F\); the family is consistent. More generally, for a finite event \(A\subseteq\{0,1\}^F\), summing the point masses over assignments on added coordinates gives \(\mu_G(\pi_{G,F}^{-1}(A))=\mu_F(A)\).
The event \(C=\{x:x_2=1,\ x_5=0\}\) is represented on \(F=\{2,5\}\) by the single point \((1,0)\), whose mass is \((1/3)^1(2/3)^0(1/3)^0(2/3)^1=2/9\). If instead we represent it on \(G=\{1,2,5\}\), both values of coordinate 1 are allowed, so the probability is \((2/3)(2/9)+(1/3)(2/9)=4/27+2/27=2/9\). Check: the two representations give the same probability, as consistency requires.
Worked Example: A Dependent Consistent Family
For each nonempty finite \(F\), put mass \(2/3\) on the all-zero point of \(\{0,1\}^F\) and mass \(1/3\) on the all-one point, with zero mass elsewhere. Let \(\mu_{\varnothing}\) be the unit mass on \(S_{\varnothing}\). These laws are consistent: projecting either constant point to a smaller coordinate set gives the corresponding constant point, so the two masses remain \(2/3\) and \(1/3\).
For \(C=\{x:x_1=1\text{ or }x_2=1\}\), the all-zero point is excluded and the all-one point is included, so \(\mu_0(C)=1/3\). For \(D=\{x:x_1=1,\ x_2=0\}\), both constant points are excluded, so \(\mu_0(D)=0\). In particular the two coordinates are not independent under these finite laws: each has probability \(1/3\) of being 1, but the probability that both equal 1 is \(1/3\), not \((1/3)(1/3)=1/9\). This example shows that consistency alone does not imply independence.
Check: on the two-point finite product, the four masses are \(2/3\) at \((0,0)\), \(1/3\) at \((1,1)\), and 0 at each of \((0,1)\) and \((1,0)\); they sum to \(2/3+1/3+0+0=1\).
Consistency Is What Makes the Assignment Work
A common mistake is to assign a cylinder probability using whichever finite-dimensional law is most convenient, without checking consistency. For a concrete failure, let \(\mu_{\{1\}}\) give probability \(1/2\) to coordinate value 1. Let \(\mu_{\{1,2\}}\) assign mass \(3/4\) to \((1,0)\), mass \(1/4\) to \((0,0)\), and zero to the other two points. The cylinder \(C=\{x:x_1=1\}\) can be represented on \(\{1\}\), giving the proposed value \(1/2\). Representing it on \(\{1,2\}\) gives \(3/4\). The two assignments disagree because the marginal of \(\mu_{\{1,2\}}\) on coordinate 1 gives probability \(3/4\) to 1, not \(1/2\).
The correction is to check the marginalization identity for every inclusion of finite coordinate sets before using the family to assign cylinder probabilities. Once consistency holds, the common-coordinate argument in the theorem proves that all representations agree. The algebra and its finitely additive probability are useful starting data for constructing a measure on the infinite product, but the construction and the additional countable-additivity issues require further results. In particular, finite additivity on this algebra should not be silently treated as a completed probability measure on the product sigma-algebra.
Check Your Understanding
Use the definitions and the common-coordinate method to answer these questions.
- Write the complement of \(C(F,A)\) as a cylinder on the same finite coordinate set.
- Explain why the union of two cylinders depending on different finite sets can be represented on their union of coordinate sets.
- In the Bernoulli family with probability \(1/3\) of value 1, compute the probability of \(x_1=0,\ x_4=1,\ x_6=1\).
- Why does finite additivity of cylinder probabilities not by itself establish a probability measure on the product sigma-algebra?
- Give a reason the set of sequences with at least one coordinate equal to 1 is not a cylinder in \(\{0,1\}^{\mathbb{N}}\).