Two Sigma-Algebras on a Product
When each factor measure space is complete, it is tempting to expect that the product measure space is complete as well. That conclusion need not hold. The key distinction is between the sigma-algebra generated by measurable rectangles and the completion of the measure on that sigma-algebra. The completion adds every subset of every product-measurable null set; the product sigma-algebra itself need not contain those subsets.
For measurable spaces \((X,\mathcal{M})\) and \((Y,\mathcal{N})\), the product sigma-algebra is \(\mathcal{M}\otimes\mathcal{N}=\sigma(\{A\times B:A\in\mathcal{M},\,B\in\mathcal{N}\})\). If \(\mu\) and \(\nu\) are sigma-finite measures, their product measure \(\mu\times\nu\) is defined on \(\mathcal{M}\otimes\mathcal{N}\) and satisfies \((\mu\times\nu)(A\times B)=\mu(A)\nu(B)\). A measure space is complete when every subset of each measurable null set is measurable.
We will use the finite product measure \(\mu=m|_{\mathcal{L}_{[0,1]}}\times m|_{\mathcal{L}_{[0,1]}}\) on \(([0,1]^2,\mathcal{L}_{[0,1]}\otimes\mathcal{L}_{[0,1]})\). Here \(\mathcal{L}_{[0,1]}\) is the Lebesgue sigma-algebra restricted to \([0,1]\), and \(m|_{\mathcal{L}_{[0,1]}}\) is Lebesgue measure restricted to it. Each factor has total measure \(1\), so it is finite, hence sigma-finite; the product measure is finite and \(\mu([0,1]^2)=1\). Each factor is complete, but we will see that this product measure space is not.
This definition is well-defined: if \(E\triangle A\subseteq N\) and \(E\triangle B\subseteq N'\) with \(N,N'\) null and measurable, then \(A\triangle B\subseteq N\cup N'\), so \(\mu(A)=\mu(B)\). Also, \(\overline{\mathcal{M}}\) is a sigma-algebra. Complements preserve the symmetric-difference condition, and for \(E_j\triangle A_j\subseteq N_j\), the union \(\bigcup_j E_j\) differs from \(\bigcup_j A_j\) only within the measurable null set \(\bigcup_jN_j\).
Proof. The formula \(\overline{\mu}(E)=\mu(A)\) is countably additive. To verify this, take pairwise disjoint \(E_j\in\overline{\mathcal{M}}\) and measurable approximants \(A_j\) with \(E_j\triangle A_j\subseteq N_j\), where each \(N_j\) is measurable and null. Define the disjoint measurable sets \(A'_1=A_1\) and \(A'_j=A_j\setminus\bigcup_{i<j}A_i\) for \(j\ge2\). For each \(j\), \(A'_j\triangle E_j\) is contained in \(\bigcup_{i\le j}N_i\): outside that union, membership in each \(A_i\) agrees with membership in \(E_i\), and the \(E_i\) are disjoint. Thus \(\overline{\mu}(E_j)=\mu(A'_j)\). Further, \(\bigcup_j A'_j=\bigcup_j A_j\) differs from \(\bigcup_j E_j\) only within \(\bigcup_jN_j\). Therefore $$ \overline{\mu}\Bigl(\bigcup_j E_j\Bigr)=\mu\Bigl(\bigcup_jA'_j\Bigr)=\sum_j\mu(A'_j)=\sum_j\overline{\mu}(E_j), $$ where the middle equality is countable additivity of \(\mu\) on the disjoint measurable sets \(A'_j\). Finally, if \(\overline{\mu}(E)=0\), an approximant \(A\) can be chosen with \(\mu(A)=0\), and \(E\subseteq A\cup N\) for a measurable null set \(N\). Every subset of \(E\) then belongs to \(\overline{\mathcal{M}}\) by the defining condition. Thus the completed measure is complete. \(\square\)
A Null Set That Contains More Than the Product Sigma-Algebra
First construct a non-Lebesgue-measurable set \(T\subseteq[0,1]\). Declare \(x,y\in[0,1]\) equivalent when \(x-y\in\mathbb{Q}\), and choose one representative from each equivalence class. The resulting set \(T\) is a Vitali set. Its nonmeasurability follows directly from translation invariance and countable additivity.
Proof. The sets \(T+q\), for \(q\in\mathbb{Q}\cap[-1,1]\), are pairwise disjoint. Indeed, if \(t_1+q_1=t_2+q_2\), then \(t_1-t_2=q_2-q_1\in\mathbb{Q}\), so the representatives \(t_1,t_2\) belong to the same equivalence class and must be equal; consequently \(q_1=q_2\). Their union covers \([0,1]\), since every \(x\in[0,1]\) differs rationally from its chosen representative \(t\in T\), and \(x=t+q\) for some \(q\in\mathbb{Q}\cap[-1,1]\). All these translates lie in \([-1,2]\). If \(T\) were measurable with measure zero, their countable union would have measure zero, contradicting that it contains \([0,1]\), whose measure is \(1\). If \(T\) were measurable with positive measure, countably infinitely many disjoint translates of \(T\), all with the same positive measure, would have infinite total measure inside \([-1,2]\), whose measure is \(3\). Both possibilities are impossible, so \(T\) is not Lebesgue measurable. \(\square\)
Worked Example: A Null Strip and Its Arbitrary Subsets
Let \(C\subseteq[0,1]\) be the Cantor set and \(E=C\times[0,1]\). Determine its product measure and locate every subset \(H\subseteq E\) in the product space and its completion.
The Cantor set is Borel, hence belongs to \(\mathcal{L}_{[0,1]}\). At stage \(n\) of its construction it is covered by \(2^n\) closed intervals, each of length \(3^{-n}\). The total length of that cover is \(2^n3^{-n}=(2/3)^n\), which tends to zero. Hence \(m(C)=0\). Since \(E\) is a measurable rectangle, the rectangle formula gives $$ \mu(E)=m(C)m([0,1])=0\cdot1=0. $$ Thus \(E\in\mathcal{L}_{[0,1]}\otimes\mathcal{L}_{[0,1]}\), and every subset \(H\subseteq E\) belongs to the completion, with completed measure zero. The value is checked by the same rectangle formula: the measure of the containing rectangle is already zero. The argument does not imply that every such \(H\) belongs to the product sigma-algebra; the completion is precisely where all these subsets are guaranteed to be included.
Worked Example: A Nonmeasurable Subset of a Null Line
Consider \(H=T\times\{0\}\subseteq[0,1]^2\), where \(T\) is the Vitali set above. Show that \(H\) belongs to the completion but not to the product sigma-algebra.
The containing line \(L=[0,1]\times\{0\}\) is a measurable rectangle, and $$ \mu(L)=m([0,1])m(\{0\})=1\cdot0=0. $$ Therefore \(H\) belongs to the completion and has completed measure zero. To test product measurability, define \(j:[0,1]\to[0,1]^2\) by \(j(x)=(x,0)\). For every generating rectangle \(A\times B\), with \(A,B\in\mathcal{L}_{[0,1]}\), $$ j^{-1}(A\times B)= \begin{cases} A,&0\in B,\\ \varnothing,&0\notin B. \end{cases} $$ Both possibilities are Lebesgue measurable. Since inverse images preserve complements and countable unions, \(j^{-1}(G)\) is Lebesgue measurable for every \(G\) in the product sigma-algebra. But \(j^{-1}(H)=T\), which is not Lebesgue measurable. Thus \(H\) is not product measurable. As a check, its inclusion \(H\subseteq L\) still guarantees completed measure zero; nonmeasurability in the product sigma-algebra does not prevent membership in the completion.
The Diagonal Gives the Same Strict Inclusion
The diagonal provides a particularly direct witness. Write \(D=\{(x,x):x\in[0,1]\}\) and \(S=\{(x,x):x\in T\}\). The diagonal \(D\) is product measurable: it is closed in the square, and every relatively open subset of \([0,1]^2\) is a countable union of relatively open rectangles from a countable base. Such rectangles belong to the product sigma-algebra, so Borel subsets of the square, including \(D\), belong to it.
Worked Example: A Nonmeasurable Subset of the Null Diagonal
Show that \(\mu(D)=0\), that \(S\) belongs to the completion, and that \(S\) does not belong to the product sigma-algebra.
For each positive integer \(n\), cover \(D\) by the \(n\) rectangles $$ [k/n,(k+1)/n]\times[k/n,(k+1)/n],\qquad k=0,\ldots,n-1. $$ Each rectangle has measure \((1/n)(1/n)=1/n^2\); therefore subadditivity gives \(\mu(D)\le n(1/n^2)=1/n\). Since this holds for every positive integer \(n\) and \(\mu(D)\ge0\), \(\mu(D)=0\). As \(S\subseteq D\), the completion contains \(S\) and assigns it measure zero.
Now define \(d:[0,1]\to[0,1]^2\) by \(d(x)=(x,x)\). On a generating rectangle, $$ d^{-1}(A\times B)=A\cap B\in\mathcal{L}_{[0,1]}. $$ Inverse images preserve sigma-algebra operations, so \(d^{-1}(G)\) is Lebesgue measurable for every product-measurable \(G\). But \(d^{-1}(S)=T\), which is not Lebesgue measurable. Hence \(S\) is not in the product sigma-algebra. For a numerical check of the cover estimate, when \(n=4\) the four squares each have measure \(1/16\), giving total measure \(4/16=1/4\); as \(n\) increases, the bound \(1/n\) tends to zero, as required.
Proof. Every product-measurable set belongs to the completion: in the completion definition take \(A=E\) and \(N=\varnothing\). Thus the product sigma-algebra is contained in its completion. The set \(S\) constructed above belongs to the completion because it is a subset of the product-measurable null set \(D\), but it does not belong to the product sigma-algebra. The containment is therefore strict. \(\square\)
What Completeness Does—and Does Not—Say
Completeness of each factor means that subsets of factor-measurable null sets are factor measurable. It does not say that every subset of a product-measurable null set can be built from measurable rectangles using sigma-algebra operations. In the examples, the product sigma-algebra contains the null sets \(L\) and \(D\), while the completion adds subsets that are not product measurable.
A common mistake is to argue, “The diagonal is measurable and null, so every subset of it is product measurable.” That conclusion applies only when the measure space is complete. Here it is the completion, not the original product sigma-algebra, that guarantees measurability of every subset of \(D\). The diagonal-map test makes the failure concrete: if the particular subset \(S\) were product measurable, its inverse image under \(d\) would be Lebesgue measurable, forcing the impossible conclusion that \(T\) is Lebesgue measurable.
Check Your Understanding
Use the definitions and the distinction between the product sigma-algebra and its completion to answer the following.
- What is the product measure of \(C\times[0,1]\), and why does every subset of this rectangle belong to the completion?
- Why does completeness of both factor Lebesgue spaces not imply completeness of their product?
- For the diagonal map \(d(x)=(x,x)\), compute \(d^{-1}(A\times B)\) and explain why this verifies measurability of inverse images of all product-measurable sets.
- Why does the fact that \(S\subseteq D\) establish membership in the completion but not membership in the product sigma-algebra?
- In the diagonal covering argument, what bound is obtained using \(n\) squares, and why does that bound imply \(\mu(D)=0\)?