Why Sections Preserve Measurability
For a set \(E\subseteq X\times Y\), fixing one coordinate leaves a subset of the other space. Such sections are useful because they turn a question about a set in a product space into a question about sets in a factor space. Rectangles have measurable sections immediately. The substantive point is that this property persists under the countable operations that generate a product sigma-algebra.
Let \((X,\mathcal{M})\) and \((Y,\mathcal{N})\) be measurable spaces. Their product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) is the sigma-algebra on \(X\times Y\) generated by the measurable rectangles \(A\times B\), where \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). No measure, finiteness, sigma-finiteness, or completeness assumption is needed for the results below.
For a rectangle, the section formulas are \( (A\times B)_x=B\) if \(x\in A\), and \(\varnothing\) otherwise; and \( (A\times B)^y=A\) if \(y\in B\), and \(\varnothing\) otherwise. Thus both sections belong to their respective factor sigma-algebras. To extend this fact from rectangles to all product-measurable sets, we use a pi-lambda argument.
The Pi-Lambda Argument
Proof of the pi-lambda theorem. Let \(\mathcal{D}_0\) be the smallest lambda-system containing \(\mathcal{P}\), obtained by intersecting all such lambda-systems. A lambda-system is closed under differences \(B\setminus A\) when \(A\subseteq B\) and both sets belong to it: \(B^c\) and \(A\) are disjoint members, so \(B^c\cup A\) belongs to it, and its complement is \(B\setminus A\).
Fix \(A\in\mathcal{P}\). The family \(\mathcal{C}_A=\{B\in\mathcal{D}_0:A\cap B\in\mathcal{D}_0\}\) is a lambda-system. It contains \(S\) because \(A\in\mathcal{D}_0\). If \(B\in\mathcal{C}_A\), then \(A\cap B^c=A\setminus(A\cap B)\in\mathcal{D}_0\) by the nested-difference property. If the \(B_j\) are pairwise disjoint members of \(\mathcal{C}_A\), then the sets \(A\cap B_j\) are pairwise disjoint members of \(\mathcal{D}_0\), so their union belongs to \(\mathcal{D}_0\). Also \(\mathcal{P}\subseteq\mathcal{C}_A\), since \(\mathcal{P}\) is closed under intersections. Minimality of \(\mathcal{D}_0\) gives \(\mathcal{D}_0\subseteq\mathcal{C}_A\). Thus \(A\cap B\in\mathcal{D}_0\) for every \(A\in\mathcal{P}\) and \(B\in\mathcal{D}_0\).
Now fix \(B\in\mathcal{D}_0\) and consider \(\mathcal{C}_B=\{A\in\mathcal{D}_0:A\cap B\in\mathcal{D}_0\}\). The same closure checks show that \(\mathcal{C}_B\) is a lambda-system. The preceding conclusion shows it contains \(\mathcal{P}\); hence it contains \(\mathcal{D}_0\). Therefore \(\mathcal{D}_0\) is closed under finite intersections. A lambda-system closed under finite intersections is a sigma-algebra: it is closed under complements, and any countable union can be written as a disjoint union by removing from each set the preceding finite union. Consequently \(\mathcal{D}_0\) is a sigma-algebra containing \(\mathcal{P}\), so \(\sigma(\mathcal{P})\subseteq\mathcal{D}_0\subseteq\mathcal{D}\). \(\square\)
Proof. Fix \(y\in Y\), and define \(\mathcal{D}_y=\{E\subseteq X\times Y:E^y\in\mathcal{M}\}\). This is a lambda-system on \(X\times Y\). Indeed, \((X\times Y)^y=X\in\mathcal{M}\); \((E^c)^y=X\setminus E^y\); and if the sets \(E_j\) are pairwise disjoint, then their sections \(E_j^y\) are pairwise disjoint and \((\bigcup_j E_j)^y=\bigcup_j E_j^y\). Since \(\mathcal{M}\) is a sigma-algebra, these operations preserve membership in \(\mathcal{M}\). Every measurable rectangle belongs to \(\mathcal{D}_y\), by the section formula for rectangles, including the case \(y\notin B\), when the section is empty. Measurable rectangles form a pi-system containing \(X\times Y\), and they generate \(\mathcal{M}\otimes\mathcal{N}\). The pi-lambda theorem gives \(\mathcal{M}\otimes\mathcal{N}\subseteq\mathcal{D}_y\). Since \(y\) was arbitrary, every horizontal section is measurable. Fixing \(x\in X\) instead and using \(\{E:E_x\in\mathcal{N}\}\) gives, by the same lambda-system checks and the rectangle formula, that every vertical section is measurable. \(\square\)
Worked Example: Sections of a Rectangle
Take \(X=Y=\mathbb{R}\) with \(\mathcal{M}=\mathcal{N}=\mathcal{B}_{\mathbb{R}}\), the Borel sigma-algebra, and let \(E=[0,2)\times(1,4]\). Find the horizontal section at \(y=3\) and the vertical section at \(x=2\).
For the horizontal section, the fixed coordinate is \(y=3\). Since \(3\in(1,4]\), the rectangle formula gives \(E^3=[0,2)\). For the vertical section at \(x=2\), membership requires \(2\in[0,2)\). This is false because the right endpoint is excluded, so \(E_2=\varnothing\). Both \([0,2)\) and \(\varnothing\) are Borel sets. Check: the point \((1,3)\) belongs to \(E\), so \(1\in E^3\); the point \((2,3)\) does not, so \(3\notin E_2\). The endpoint tests agree with the computed sections.
Worked Example: Sections of a Closed Region
In \(\mathbb{R}^2\) with the Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\), let \(E=\{(x,y):x\le y\}\). Show that \(E\) is product-measurable and compute its sections at \(y=2\) and \(x=2\).
The map \((x,y)\mapsto x-y\) is continuous, and \(E\) is its inverse image of the closed set \((-\infty,0]\). Hence \(E\) is closed in \(\mathbb{R}^2\), and therefore belongs to the Borel sigma-algebra on \(\mathbb{R}^2\), which is \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\). At \(y=2\), the condition defining \(E^2\) is \(x\le2\), so \(E^2=(-\infty,2]\). At \(x=2\), the condition defining \(E_2\) is \(2\le y\), so \(E_2=[2,\infty)\). These are Borel sets. Check: \((2,2)\in E\), so \(2\) belongs to both sections; \((3,2)\notin E\), so \(3\notin E^2\); and \((2,1)\notin E\), so \(1\notin E_2\). The boundary is included because the defining inequality is non-strict.
Sections of Product-Measurable Functions
Proof. First suppose \(f\) is real-valued and fix \(x\in X\). For any Borel set \(C\subseteq\mathbb{R}\), the set \(E=f^{-1}(C)\) belongs to \(\mathcal{M}\otimes\mathcal{N}\) by measurability of \(f\). Directly from the definitions, \((f_x)^{-1}(C)=\{y:f(x,y)\in C\}=(f^{-1}(C))_x=E_x\). The set-section theorem gives \(E_x\in\mathcal{N}\). Thus the inverse image under \(f_x\) of every Borel set is in \(\mathcal{N}\), which is exactly measurability of \(f_x\). For an extended-real or complex-valued function, use the same identity with Borel sets in its stated codomain; the set-section theorem applies without change. Fixing \(y\) instead gives the horizontal-section conclusion. These identities and conclusions hold for every fixed coordinate. \(\square\)
Worked Example: Sections of a Product-Measurable Simple Function
Let \(X=Y=\mathbb{R}\), both with their Borel sigma-algebras, and define \(f=2\chi_{[0,1]\times(1,3]}+3\chi_{(2,4)\times\{0\}}\). The two rectangles are disjoint because their second-coordinate sets are disjoint. Compute \(f_x\) for all \(x\), and verify its measurability.
Both rectangles belong to \(\mathcal{B}_{\mathbb{R}}\otimes\mathcal{B}_{\mathbb{R}}\), so their indicators and the stated finite linear combination are product-measurable. If \(x\in[0,1]\), the first rectangle contributes \(2\) exactly when \(y\in(1,3]\), while the second contributes nothing. Hence \(f_x=2\chi_{(1,3]}\). If \(x\in(2,4)\), the first rectangle contributes nothing and the second contributes \(3\) exactly at \(y=0\), so \(f_x=3\chi_{\{0\}}\). For every other \(x\), neither rectangle contains a point with first coordinate \(x\), and \(f_x\) is the zero function. In particular, \(x=1\) is in the first case, while \(x=2\) is in the third case because the interval \((2,4)\) excludes \(2\).
Each section is Borel-measurable: for example, \((1,3]\) and \(\{0\}\) are Borel sets, and the zero function is measurable. To check the values directly, \(f_1(2)=2\), \(f_3(0)=3\), and \(f_2(2)=0\). These follow from the indicated endpoint conventions and the original formula. As a further inverse-image check, \((f_1)^{-1}(\{2\})=(1,3]\), \((f_3)^{-1}(\{3\})=\{0\}\), and \((f_2)^{-1}(\{2\})=\varnothing\); all three are Borel sets.
Scope of the Result and a Common Pitfall
The theorem concerns the product sigma-algebra itself. It does not assert that every section of every set in the completion of a product measure is measurable in the factor sigma-algebra. For example, let \(m\) be Lebesgue measure on \(\mathbb{R}\), and choose a non-Lebesgue-measurable set \(V\subseteq[0,1]\). The line \(\mathbb{R}\times\{0\}\) is in \(\mathcal{L}\otimes\mathcal{L}\) and is null: it is the countable union of the rectangles \([-n,n]\times\{0\}\), each of product measure \(m([-n,n])m(\{0\})=0\). Thus \(V\times\{0\}\), being a subset of this null set, belongs to the completion of the product measure. But its horizontal section at \(y=0\) is \((V\times\{0\})^0=V\), which is not in \(\mathcal{L}\). This does not contradict the theorem: \(V\times\{0\}\) need not belong to the uncompleted product sigma-algebra \(\mathcal{L}\otimes\mathcal{L}\).
A related notation error is to call \(E^y\) the vertical section at \(y\). By definition, \(E^y\) fixes the second coordinate and leaves a subset of \(X\), so it is the horizontal section. The vertical section fixes \(x\) and is \(E_x\subseteq Y\). In the example above the nonmeasurable section is horizontal, at \(y=0\), not vertical.
Check Your Understanding
For each question, identify which coordinate is fixed and which factor sigma-algebra must contain the resulting section.
- For \(E=A\times B\), derive both \(E_x\) and \(E^y\), including the cases \(x\notin A\) and \(y\notin B\).
- Explain why \(\{E:E^y\in\mathcal{M}\}\) is a lambda-system and why it contains all measurable rectangles.
- Let \(E=\{(x,y):x<y\}\) in the Borel product on \(\mathbb{R}^2\). Compute \(E^0\) and \(E_0\), and state whether either endpoint is included.
- For a product-measurable \(f:X\times Y\to\mathbb{R}\), write the identity relating \((f_x)^{-1}(C)\) to a section of \(f^{-1}(C)\), for Borel \(C\subseteq\mathbb{R}\).
- Why does the completed-product example with \(V\times\{0\}\) not contradict measurability of sections for sets in \(\mathcal{L}\otimes\mathcal{L}\)?