Finite-Coordinate Information in a Sequence Space
A point in a countable product is a sequence, but a measurable event need not be determined by any fixed finite list of its coordinates. For example, the property that a sequence visits a set infinitely often asks about every tail of the sequence. Cylinders provide the basic measurable events that inspect finitely many coordinates; countable sigma-algebra operations then build events involving the whole sequence.
Let \((X_n,\mathcal{M}_n)\), \(n\in\mathbb{N}=\{1,2,\ldots\}\), be measurable spaces, and let \(X=\prod_{n=1}^{\infty}X_n\) be their Cartesian product. A point \(x\in X\) is written \(x=(x_1,x_2,\ldots)\). We do not assume that any measure has been specified on these spaces. The product sigma-algebra is the sigma-algebra generated by the coordinate maps \(\pi_n:X\to X_n\), where \(\pi_n(x)=x_n\); explicitly, it is \(\sigma(\{\pi_n^{-1}(A): n\in\mathbb{N},\,A\in\mathcal{M}_n\})\), denoted \(\bigotimes_{n=1}^{\infty}\mathcal{M}_n\).
A cylinder need not be a product of separate conditions on each coordinate: \(B\) can express a relation among the coordinates in \(F\). For instance, in \(\mathbb{R}^{\mathbb{N}}\) with each factor equipped with \(\mathcal{B}_{\mathbb{R}}\), the set \(\{x:x_2<x_5\}\) is a cylinder based on coordinates \(2\) and \(5\). Its defining set in \(\mathbb{R}^2\) is \(\{(u,v):u<v\}\), a Borel set.
Why Cylinders Generate the Product Sigma-Algebra
Proof. Denote the product sigma-algebra by \(\mathcal{M}\), and let \(\mathcal{C}\) be the collection of all finite-coordinate cylinders. First, every \(C=\pi_F^{-1}(B)\in\mathcal{C}\) belongs to \(\mathcal{M}\). To verify this, consider the collection \(\mathcal{D}\) of sets \(B\in\mathcal{M}_F\) for which \(\pi_F^{-1}(B)\in\mathcal{M}\). In the finite product \(X_F\), measurable rectangles \(\prod_{i\in F}A_i\), with \(A_i\in\mathcal{M}_i\), generate \(\mathcal{M}_F\). For each such rectangle, \(\pi_F^{-1}(\prod_{i\in F}A_i)=\bigcap_{i\in F}\pi_i^{-1}(A_i)\in\mathcal{M}\), since \(F\) is finite and each coordinate inverse image is a generator of \(\mathcal{M}\). The collection \(\mathcal{D}\) is a sigma-algebra: inverse images preserve complements and countable unions. It contains the generating rectangles, so \(\mathcal{D}=\mathcal{M}_F\). Thus \(\mathcal{C}\subseteq\mathcal{M}\), and consequently \(\sigma(\mathcal{C})\subseteq\mathcal{M}\).
For the reverse inclusion, take any generator \(\pi_n^{-1}(A)\) of \(\mathcal{M}\), where \(A\in\mathcal{M}_n\). This is a finite-coordinate cylinder with \(F=\{n\}\) and \(B=A\). Therefore every generator of \(\mathcal{M}\) belongs to \(\mathcal{C}\), so \(\mathcal{M}\subseteq\sigma(\mathcal{C})\). The two inclusions give \(\mathcal{M}=\sigma(\mathcal{C})\). \(\square\)
This equality says that every product-measurable event can be assembled from finite-coordinate cylinders by taking complements and countable unions (equivalently, by countable sigma-algebra operations). It does not say that every such event is itself one cylinder. In particular, membership in a general product-measurable set may depend on infinitely many coordinates.
Events Occurring Infinitely Often or Eventually
Here is a useful consequence in a general countable product \((X,\mathcal{M})\) as above. Suppose \(A_n\in\mathcal{M}_n\) for each \(n\), and let \(C_n=\{x\in X:x_n\in A_n\}=\pi_n^{-1}(A_n)\). Each \(C_n\) is a cylinder and is in \(\mathcal{M}\). The event that the coordinate condition holds infinitely often is \(\bigcap_{N=1}^{\infty}\bigcup_{n\ge N}C_n\). The event that it holds for all sufficiently large indices is \(\bigcup_{N=1}^{\infty}\bigcap_{n\ge N}C_n\).
Proof. For each \(N\), \(\bigcup_{n\ge N}C_n\) is measurable because it is a countable union of measurable sets. Their countable intersection over \(N\) is measurable. A point \(x\) belongs to this intersection exactly when, for every \(N\), at least one \(n\ge N\) satisfies \(x_n\in A_n\). That is exactly the condition of infinitely many such indices. Also, for each \(N\), \(\bigcap_{n\ge N}C_n\) is measurable by closure under countable intersections. Their countable union is measurable. Membership means that for some \(N\), every \(n\ge N\) satisfies the condition, which is precisely “eventually always.” \(\square\)
Worked Example: Infinitely Many Visits and Eventual Membership
Take \(X=\{0,1\}^{\mathbb{N}}\), with each factor carrying the full discrete sigma-algebra, and let \(A_n=\{1\}\) for every \(n\). The coordinate event is \(C_n=\{x:x_n=1\}\). Thus the event of infinitely many visits to \(1\) is \(I=\bigcap_{N=1}^{\infty}\bigcup_{n\ge N}C_n\), and the event of eventually always being \(1\) is \(T=\bigcup_{N=1}^{\infty}\bigcap_{n\ge N}C_n\). Both are measurable by the proposition.
For \(x=(1,0,1,0,\ldots)\), every tail contains a \(1\), so \(x\in\bigcup_{n\ge N}C_n\) for every \(N\), hence \(x\in I\). But for each \(N\), the index \(n=N+1\) has \(x_n=0\) when \(N\) is odd, and the index \(n=N+2\) has \(x_n=0\) when \(N\) is even. Thus no tail consists entirely of \(1\)'s, and \(x\notin T\). For \(y=(0,1,1,1,\ldots)\), every tail contains only \(1\)'s once \(N=2\), so \(y\in T\), and therefore \(y\in I\) as well.
Check: eventual membership implies infinitely many visits here because a sequence that is \(1\) at every index \(n\ge N\) has infinitely many such indices. The alternating sequence belongs to \(I\) but not \(T\), confirming that the two events are not the same.
Measurability of Basic Sequence Properties
For real sequences, use \(X=\mathbb{R}^{\mathbb{N}}\) and equip every factor with \(\mathcal{B}_{\mathbb{R}}\). The product sigma-algebra in the following examples is \(\bigotimes_{n=1}^{\infty}\mathcal{B}_{\mathbb{R}}\); no measure on this space is assumed. Coordinate conditions involving Borel subsets of \(\mathbb{R}\), and conditions involving finitely many coordinates through a Borel subset of \(\mathbb{R}^k\), define measurable cylinders.
Worked Example: Sequences Converging to Zero
Let \(Z=\{x\in\mathbb{R}^{\mathbb{N}}:x_n\to0\}\). For each positive integer \(k\), set \(U_{k,n}=\{x:|x_n|<1/k\}\). This is a cylinder because \((-1/k,1/k)\) is open, hence Borel. The convergence criterion gives the exact expression
Indeed, if \(x_n\to0\), then for each \(k\) there is an \(N\) such that \(|x_n|<1/k\) for every \(n\ge N\); hence \(x\) belongs to the right-hand side. Conversely, suppose \(x\) belongs to that side. Given any \(\varepsilon>0\), choose \(k\in\mathbb{N}\) with \(1/k<\varepsilon\). The membership condition supplies \(N\) such that \(|x_n|<1/k<\varepsilon\) for every \(n\ge N\). This is the definition of \(x_n\to0\). The right-hand side is measurable by countable intersections and unions, so \(Z\in\bigotimes_n\mathcal{B}_{\mathbb{R}}\).
For \(x_n=1/n\), given \(k\), take \(N=k+1\); then \(n\ge N\) implies \(1/n\le1/(k+1)<1/k\). Thus this sequence belongs to \(Z\). For the constant sequence \(y_n=1\), the condition for \(k=1\) fails at every index because \(1<1\) is false, so \(y\notin Z\). Check: the strict inequality in the definition is respected: at \(n=k\), \(1/n=1/k\) would not satisfy \(1/n<1/k\), which is why \(N=k+1\) is chosen.
Worked Example: Eventually Zero Sequences
Still in \(\mathbb{R}^{\mathbb{N}}\) with the product Borel sigma-algebra, consider \(E=\{x:x_n=0\text{ for all sufficiently large }n\}\). For each \(n\), \(V_n=\{x:x_n=0\}\) is a cylinder, since \(\{0\}\) is Borel. The event is
For a fixed \(N\), the intersection requires \(x_n=0\) at every index \(n\ge N\). Taking the union over \(N\) allows any finite initial segment, so this expression is exactly the stated event. It is measurable by countable sigma-algebra operations.
The sequence \(x=(3,-2,0,0,0,\ldots)\) belongs to \(\bigcap_{n\ge3}V_n\), hence to \(E\). The sequence \(y_n=1/n\) does not belong to \(E\): for every \(N\), the index \(n=N\) satisfies \(y_n=1/N\ne0\), so \(y\notin\bigcap_{n\ge N}V_n\) for every \(N\). Check: the event requires exact equality with zero after some index; convergence to zero alone does not suffice.
A Common Pitfall: A Generating Family Is Not Usually a Sigma-Algebra
The theorem concerns the sigma-algebra generated by cylinders, not the cylinder family itself. In \(\mathbb{R}^{\mathbb{N}}\), the singleton consisting of the zero sequence is measurable, since
Each set in this intersection is a cylinder. But the singleton is not itself a finite-coordinate cylinder. If it were determined by a finite set \(F\) of coordinates, choose an index \(j\notin F\), which is possible because \(\mathbb{N}\) is infinite. The zero sequence and the sequence equal to zero except for value \(1\) at coordinate \(j\) agree on every coordinate in \(F\). A set determined only by \(F\) must either contain both sequences or contain neither. The singleton contains the zero sequence but not the altered sequence, a contradiction.
Thus countable intersections are essential: a condition may impose restrictions at every coordinate even though each individual restriction inspects only one coordinate. This distinction is useful whenever a sequence event is described in words. Translate “infinitely often” into an intersection of tail unions, “eventually always” into a union of tail intersections, and convergence into countable tests at increasingly small tolerances. The resulting expressions establish product measurability without assuming a product measure or interchanging any limits and integrals.
Check Your Understanding
For each question, work in the stated product sigma-algebra and identify the finite-coordinate events before taking countable operations.
- For measurable \(A_n\in\mathcal{M}_n\), write the event that \(x_n\in A_n\) for only finitely many indices as a countable set expression.
- In \(\mathbb{R}^{\mathbb{N}}\) with the product Borel sigma-algebra, express the event \(x_n\to a\), for a fixed real number \(a\), using strict tolerance inequalities.
- Show that \(\{x\in\mathbb{R}^{\mathbb{N}}:x_n\ge0\text{ for every }n\}\) is measurable.
- Explain why a cylinder restricting a finite set \(F\) of coordinates may still encode a relation between two coordinates in \(F\).
- Give a measurable subset of \(\mathbb{R}^{\mathbb{N}}\) that is not a finite-coordinate cylinder, and justify both claims.