Sections of Completed-Product-Measurable Sets

Completion Can Change What a Section Looks Like

For a product-measurable set, every section is measurable in the corresponding factor space. Passing to the completion of the product measure changes this pointwise conclusion: the completed space includes every subset of a product-measurable null set, and such a subset can have a nonmeasurable section. The distinction is compatible with Tonelli's theorem, which gives section measurability almost everywhere, not at every parameter.

Throughout, let $(\mathbb{R},\mathcal{L},m)$ be the complete, sigma-finite Lebesgue measure space, and let $(\mathbb{R}^2,\mathcal{L}\otimes\mathcal{L},m\times m)$ be its product measure space. The product measure space is sigma-finite, but its sigma-algebra need not be complete. Write $\widehat{\mathcal{L}\otimes\mathcal{L}}$ for the completion of $\mathcal{L}\otimes\mathcal{L}$ with respect to $m\times m$. Thus a set $E\subseteq\mathbb{R}^2$ belongs to this completion if there are $B,N\in\mathcal{L}\otimes\mathcal{L}$ with $(m\times m)(N)=0$ and $E\triangle B\subseteq N$.

Definition: For $E\subseteq\mathbb{R}^2$, its vertical section at $x\in\mathbb{R}$ is $E_x=\{y\in\mathbb{R}:(x,y)\in E\}$. Its horizontal section at $y\in\mathbb{R}$ is $E^y=\{x\in\mathbb{R}:(x,y)\in E\}$. Sections are sets; their measurability is a separate question.

Sections of Product-Measurable Sets

We first record the contrast with the completed space. Every $B\in\mathcal{L}\otimes\mathcal{L}$ has $B_x\in\mathcal{L}$ for every $x\in\mathbb{R}$. To see this, fix $x$ and consider the collection of subsets $B\subseteq\mathbb{R}^2$ for which $B_x\in\mathcal{L}$. Taking a section commutes with complements and countable unions, so this collection is a sigma-algebra. It contains every measurable rectangle $A\times D$, since its section is $D$ if $x\in A$ and the empty set otherwise. The rectangles generate $\mathcal{L}\otimes\mathcal{L}$, proving the assertion. The same argument applies to horizontal sections.

The completion, by definition, includes subsets of null sets that need not themselves belong to the product sigma-algebra. To make the difference explicit, we need a non-Lebesgue-measurable set $V\subseteq[0,1]$.

Construction: On $[0,1]$, declare $u$ and $v$ equivalent when $u-v\in\mathbb{Q}$. Choose exactly one representative from each equivalence class and call the set of representatives $V$. This selection uses the axiom of choice.

Proof that $V\notin\mathcal{L}$. Let $Q=\mathbb{Q}\cap[-1,1]$. The translates $V+q$, $q\in Q$, are pairwise disjoint. Indeed, if $v_1+q_1=v_2+q_2$, then $v_1-v_2=q_2-q_1\in\mathbb{Q}$. The representatives $v_1,v_2$ must therefore be equal, and then $q_1=q_2$. Also, every $x\in[0,1]$ belongs to one of these translates: its representative $v\in V$ satisfies $x-v\in\mathbb{Q}\cap[-1,1]$. Finally, every translate lies in $[-1,2]$, since $V\subseteq[0,1]$ and $q\in[-1,1]$.

If $V$ were Lebesgue measurable, then each translate would be measurable and have measure $m(V)$. If $m(V)=0$, their countable union would have measure zero, contradicting that it contains $[0,1]$, which has measure $1$. If $m(V)>0$, countable additivity and disjointness would give the union infinite measure, contradicting that it is contained in $[-1,2]$, which has measure $3$. These cases exhaust the possibilities because $V\subseteq[0,1]$ would imply $m(V)\leq1$ if it were measurable. Thus $V\notin\mathcal{L}$. $\square$

Worked Example: One Nonmeasurable Exceptional Section

Fix $a\in\mathbb{R}$, and set $N=\{a\}\times[0,1]$ and $E=\{a\}\times V$. Verify that $E$ is measurable in the completed product space, and find all its vertical sections.

The singleton $\{a\}$ and the interval $[0,1]$ are Lebesgue measurable, so $N\in\mathcal{L}\otimes\mathcal{L}$. The rectangle formula gives $(m\times m)(N)=m(\{a\})m([0,1])=0\cdot1=0$. Since $E\subseteq N$, the definition of completion gives $E\in\widehat{\mathcal{L}\otimes\mathcal{L}}$. Its sections are

E_x= \begin{cases} V,&x=a,\\ \varnothing,&x\ne a. \end{cases}

Thus $E_a$ is not Lebesgue measurable, while every other vertical section is measurable. For comparison, each horizontal section $E^y$ is either $\{a\}$ or $\varnothing$, both Lebesgue measurable: it is $\{a\}$ when $y\in V$ and empty when $y\notin V$. Check: $E$ is contained in the null rectangle $N$, but its section at $a$ is exactly the nonmeasurable set $V$, not a subset that has become measurable merely by lying in $[0,1]$.

Why the Exceptional Section Does Not Contradict Tonelli

The exceptional parameter in the example is a null set of parameters. More generally, completed-product measurability still guarantees measurable sections outside a null set. Here is the precise statement for the Lebesgue product space in use.

Proposition: If $E\in\widehat{\mathcal{L}\otimes\mathcal{L}}$, then $E_x\in\mathcal{L}$ for $m$-almost every $x\in\mathbb{R}$. The analogous assertion holds for horizontal sections.

Proof. By the definition of completion, choose $B,N\in\mathcal{L}\otimes\mathcal{L}$ such that $(m\times m)(N)=0$ and $E\triangle B\subseteq N$. The section $B_x$ is Lebesgue measurable for every $x$, by the product-section fact proved above. The indicator $\chi_N$ is nonnegative and product measurable. Tonelli's theorem, applied to the sigma-finite Lebesgue factor spaces, gives $$ 0=(m\times m)(N)=\int_{\mathbb{R}}m(N_x)\,dm(x). $$ The integrand $x\mapsto m(N_x)$ is measurable and nonnegative. A nonnegative measurable function with integral zero vanishes almost everywhere: for each positive integer $n$, the set where it exceeds $1/n$ has measure zero, by monotonicity of the integral; the set where it is positive is the countable union of those sets. Hence $m(N_x)=0$ for almost every $x$.

For each such $x$, taking sections in $E\triangle B\subseteq N$ gives $E_x\triangle B_x\subseteq N_x$. The Lebesgue sigma-algebra is complete, so every subset of the null measurable set $N_x$ is Lebesgue measurable. In particular, $E_x$ is measurable. Interchanging the two coordinates proves the horizontal-section assertion. $\square$

Worked Example: Uncountably Many Exceptional Parameters Can Occur

Let $C\subseteq[0,1]$ be the middle-thirds Cantor set and set $E=C\times V$. Verify that this set is in the completed product space and identify the parameters whose vertical sections are nonmeasurable.

At stage $n$ of the Cantor construction, the remaining set is a union of intervals with total length $(2/3)^n$. These stage sets decrease to $C$, and the first has finite measure. Continuity of measure from above therefore gives $m(C)=\lim_{n\to\infty}(2/3)^n=0$. The Cantor set is closed, hence Lebesgue measurable, and $[0,1]$ is measurable. Thus $C\times[0,1]\in\mathcal{L}\otimes\mathcal{L}$ and $$ (m\times m)(C\times[0,1])=m(C)m([0,1])=0\cdot1=0. $$ As $E\subseteq C\times[0,1]$, it belongs to the completed product sigma-algebra. Its sections satisfy $E_x=V$ for $x\in C$, and $E_x=\varnothing$ for $x\notin C$. Therefore the set of parameters with nonmeasurable sections is precisely $C$: it is uncountable, but has measure zero. Check: the formula for the section follows directly from membership in the Cartesian product, and $V$ is nonmeasurable by the translate argument above.

The Distinction That Prevents a False Calculation

A common mistake is to reason that every subset of a null set is measurable in the product sigma-algebra. This is not the definition of a sigma-algebra or of a measure space's completion. The subset property holds in the completion; it need not hold in the original product sigma-algebra.

Worked Example: Diagnosing the Invalid Product-Measurability Claim

For $E=\{a\}\times V$ from the first example, consider the calculation “$(m\times m)(E)=m(\{a\})m(V)=0$, so $E\in\mathcal{L}\otimes\mathcal{L}$.” Identify the error and give the valid conclusion.

The rectangle formula $(m\times m)(A\times D)=m(A)m(D)$ applies when both $A$ and $D$ are measurable. Here $V\notin\mathcal{L}$, so this formula cannot be applied to $\{a\}\times V$ as a product-measurable rectangle. Nor does the fact that $E\subseteq N$ imply $E\in\mathcal{L}\otimes\mathcal{L}$. What it does imply is $E\in\widehat{\mathcal{L}\otimes\mathcal{L}}$, because $N\in\mathcal{L}\otimes\mathcal{L}$ and $(m\times m)(N)=0$. The completed measure of $E$ is $0$: every subset of a measurable null set has measure zero in the completion. This conclusion does not make the section $E_a=V$ Lebesgue measurable. Check: the valid null-set containment is $E\subseteq\{a\}\times[0,1]$, whose product measure is $m(\{a\})m([0,1])=0\cdot1=0$; no measure of $V$ is used.

Accordingly, it is not valid to write an iterated integral of section measures as though $m(E_x)$ were defined for every $x$. At the exceptional parameter $a$, the section $E_a=V$ is not Lebesgue measurable, so $m(E_a)$ is not defined. The almost-everywhere section result avoids this problem: outside a null set the sections are measurable, and changes or failures on that exceptional set do not affect the almost-everywhere assertions in Tonelli's theorem. The distinction is between measurability in the completed product space and measurability of every individual section—not between a set having measure zero and having positive measure.

Check Your Understanding

Use the distinction between the product sigma-algebra and its completion in each question.

For $E=\{a\}\times V$, state $E_x$ for $x=a$ and for $x\ne a$. Which of these sections is nonmeasurable?
Why does $E\subseteq\{a\}\times[0,1]$ imply completed-product measurability, but not product measurability?
In the Cantor-set example, what is the set of parameters with nonmeasurable vertical sections, and what is its Lebesgue measure?
Which hypothesis allows Tonelli's theorem to show that $m(N_x)=0$ for almost every $x$ when $N\in\mathcal{L}\otimes\mathcal{L}$ and $(m\times m)(N)=0$?
Why is the expression $m(E_a)$ undefined for $E=\{a\}\times V$, even though $E$ has measure zero in the completed product space?

Sections of Completed-Product-Measurable Sets

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