Product-Null Sets and Their Sections
A set in a product space can be null even though some of its sections are large. The key qualification is “almost every”: when a product-measurable set has product measure zero, its sections have measure zero outside a null set in the base space. Tonelli’s theorem makes this precise by turning the measure of the set into the integral of its section measures.
Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, and use the product measure \(\mu\times\nu\) on \((X\times Y,\mathcal{M}\otimes\mathcal{N})\). Sigma-finite means that each space is a countable union of measurable sets of finite measure; the measures need not themselves be finite or complete. In this tutorial, \(E\) is in \(\mathcal{M}\otimes\mathcal{N}\), not merely an arbitrary subset of a product-null set. Its sections are
For such \(E\), the section measurability result and Tonelli’s theorem apply: \(E_x\in\mathcal{N}\) for every \(x\), \(E^y\in\mathcal{M}\) for every \(y\), and the section-measure functions are measurable with values in \([0,\infty]\). In particular, applying Tonelli to the nonnegative measurable function \(\chi_E\) gives
All three quantities may initially be infinite. The result below concerns when they are zero. The integral identity is being used only for \(\chi_E\), a nonnegative measurable function, so Tonelli applies without any assumption of integrability.
The Almost-Everywhere Section Criterion
We use the following elementary fact about nonnegative measurable functions.
Proof. If \(h=0\) outside a measurable null set \(N\), then \(h\chi_{N^c}=0\) and \(h\chi_N\) has integral zero because it is supported on a null set. Thus \(\int h\,d\mu=0\). Conversely, suppose \(\int h\,d\mu=0\). For each integer \(n\ge1\), the measurable set \(A_n=\{x:h(x)>1/n\}\) satisfies \(h\ge (1/n)\chi_{A_n}\), so monotonicity of the nonnegative integral gives \(0\le \mu(A_n)/n\le\int h\,d\mu=0\). Hence \(\mu(A_n)=0\). Since \(\{h>0\}=\bigcup_{n\ge1}A_n\), countable subadditivity implies \(\mu(\{h>0\})=0\), which is exactly \(h=0\) \(\mu\)-almost everywhere. \(\square\)
Proof. Define \(h(x)=\nu(E_x)\). Section measurability gives a measurable function \(h:X\to[0,\infty]\), and Tonelli’s theorem gives \(\int_X h\,d\mu=(\mu\times\nu)(E)\). By the lemma, this integral is zero if and only if \(h=0\) \(\mu\)-almost everywhere. This proves both directions of the first equivalence. Applying the same argument to \(k(y)=\mu(E^y)\), using the other Tonelli identity, proves both directions of the second equivalence. \(\square\)
The two assertions concern different directions through the product. The first fixes \(x\) and measures the section in \(Y\); its exceptional set lies in \(X\). The second fixes \(y\) and measures the section in \(X\); its exceptional set lies in \(Y\). Both follow from Tonelli, and neither says that every section must be null.
Worked Sections and Exceptional Fibers
Worked Example: One Exceptional Section of Positive Measure
Let \(X=Y=\mathbb{R}\) with Lebesgue measure \(m\), and take \(E=\{0\}\times[0,1]\). This is a Borel rectangle, so it is measurable for the product sigma-algebra. Its vertical sections are \(E_0=[0,1]\) and \(E_x=\varnothing\) for every \(x\ne0\). Therefore \(\nu(E_0)=m([0,1])=1\), while \(\nu(E_x)=m(\varnothing)=0\) for \(x\ne0\). The exceptional base set is exactly \(\{0\}\), whose measure is zero.
The rectangle formula gives an independent direct computation: \((m\times m)(E)=m(\{0\})m([0,1])=0\cdot1=0\). The section integral checks the same value: \(\int_{\mathbb R}m(E_x)\,dm(x)=1\cdot m(\{0\})=0\). Thus the section at \(x=0\) has positive measure, but it does not contradict the theorem because it occurs over a null set of base points.
Worked Example: A Countable Set of Exceptional Sections
Take \(X=Y=\mathbb{R}\) with Lebesgue measure and \(E=(\mathbb{Q}\cap[0,1])\times[0,1]\). Since \(\mathbb{Q}\cap[0,1]\) is countable and Borel, \(E\) is a countable union of Borel rectangles and is product-measurable. For \(x\in\mathbb{Q}\cap[0,1]\), \(E_x=[0,1]\), so \(m(E_x)=1\). For every other \(x\), \(E_x=\varnothing\), so \(m(E_x)=0\). The exceptional base set is \(\mathbb{Q}\cap[0,1]\), which is countable and hence has Lebesgue measure zero.
Consequently, the section integral is \(\int_{\mathbb R}\chi_{\mathbb{Q}\cap[0,1]}(x)\,dm(x)=m(\mathbb{Q}\cap[0,1])=0\), and Tonelli gives \((m\times m)(E)=0\). This also follows directly by writing \(E\) as the countable union of the rectangles \(\{q\}\times[0,1]\), each of product measure \(m(\{q\})m([0,1])=0\). For the other direction of sections, if \(y\in[0,1]\) then \(E^y=\mathbb{Q}\cap[0,1]\), of measure zero; if \(y\notin[0,1]\) then \(E^y=\varnothing\). Thus all horizontal sections are null, even though the vertical sections over rational \(x\) have measure one.
Worked Example: The Diagonal Has Null Sections
Let \(E=\{(x,y)\in[0,1]^2:x=y\}\) in \([0,1]^2\) with the product of Lebesgue measures. The diagonal is closed, so it is measurable. For each \(x\in[0,1]\), \(E_x=\{x\}\), and \(m(\{x\})=0\). Tonelli therefore gives \((m\times m)(E)=\int_{[0,1]}0\,dm(x)=0\).
Here is an independent covering check. Divide \([0,1]\) into \(n\) closed intervals of length \(1/n\). The \(n\) squares formed by taking each interval times itself cover the diagonal; each square has product measure \(1/n^2\). Subadditivity yields \(0\le(m\times m)(E)\le n(1/n^2)=1/n\) for every \(n\ge1\). Given any \(\varepsilon>0\), choose an integer \(n>1/\varepsilon\); then the product measure is at most \(1/n<\varepsilon\). Since it is nonnegative and this holds for every \(\varepsilon>0\), it is zero, in agreement with Tonelli.
What the Conclusion Does—and Does Not—Say
The criterion is useful because it converts a two-dimensional nullity question into a one-dimensional almost-everywhere statement about sections. It is also a precise safeguard against a common overstatement: “the product set is null, so all its sections are null” is false. In the first example, the section at \(x=0\) has measure one. The correct conclusion is that the set of such exceptional \(x\)'s is null.
The converse has the same almost-everywhere qualification. It is enough that \(\nu(E_x)=0\) for \(\mu\)-almost every \(x\); section measures at the exceptional \(x\)'s may be positive or infinite. They do not contribute to the integral over a null set. For example, in \(\{0\}\times\mathbb{R}\), the section at \(x=0\) has infinite Lebesgue measure, while all other vertical sections are empty. The product set is nevertheless null: it is contained in \([-n,n]\)-sized vertical rectangles \(\{0\}\times[-n,n]\), each of product measure \(m(\{0\})m([-n,n])=0\), and their countable union is the full set.
There is a measurability boundary to keep in view. The theorem as stated assumes \(E\in\mathcal{M}\otimes\mathcal{N}\), so the sections and their measures are measurable in the form required by Tonelli. One should not silently replace this with “\(E\) is any subset of a product-measurable null set”: such a set need not belong to the product sigma-algebra, and the section-measurability conclusion then needs separate justification. Product sigma-algebras and their completions are distinct notions.
Check Your Understanding
For each question, identify the base space for the exceptional set and use the correct direction of the section identity.
- For \(E=\{0\}\times[0,1]\subset\mathbb{R}^2\), determine \(m(E_x)\) for every \(x\), and identify the exceptional base set.
- Suppose \(E\in\mathcal{M}\otimes\mathcal{N}\) and \(\nu(E_x)=0\) for \(\mu\)-almost every \(x\). Which Tonelli identity proves that \((\mu\times\nu)(E)=0\)?
- For \(E=(\mathbb{Q}\cap[0,1])\times[0,1]\), which vertical sections have measure one, and why does this not contradict the null-section theorem?
- State the horizontal-section conclusion when \((\mu\times\nu)(E)=0\). In which space does its exceptional set lie?
- Why is it important to state that \(E\) belongs to \(\mathcal{M}\otimes\mathcal{N}\) when applying the section criterion?