Sections Turn Set Membership into an Indicator
For a set in a product space, fixing one coordinate leaves a set in the other coordinate. Integrating the indicator of that section measures how large the remaining slice is. This pointwise calculation is a basic step toward product-integration results, but it does not by itself justify integrating those slice measures over the fixed coordinate.
Throughout, let \((X,\mathcal M,\mu)\) and \((Y,\mathcal N,\nu)\) be sigma-finite measure spaces; they need not be finite or complete. The product measure \(\mu\times\nu\) is taken on \(\mathcal M\otimes\mathcal N\), not on a completion of that sigma-algebra. Let \(E\in\mathcal M\otimes\mathcal N\). For \(x\in X\), define its section by \(E_x=\{y\in Y:(x,y)\in E\}\). The measurability-of-sections result says that \(E_x\in\mathcal N\) for every \(x\). The indicator functions here are measurable, real-valued functions taking values in \(\{0,1\}\); their integrals are nonnegative extended-real integrals and may equal \(+\infty\).
Proof. Fix \(x\in X\). For each \(y\in Y\), the definition of \(E_x\) gives \((x,y)\in E\) if and only if \(y\in E_x\). Therefore \(\chi_E(x,y)=\chi_{E_x}(y)\) for every \(y\). The section \(E_x\) is in \(\mathcal N\), so its indicator is measurable. By the definition of the nonnegative Lebesgue integral of an indicator, \(\int_Y\chi_{E_x}\,d\nu=\nu(E_x)\), including when this measure is infinite. Substituting the pointwise identity proves the formula. \(\square\)
The proposition concerns integration in \(y\) with \(x\) held fixed. It does not assert that the function \(x\mapsto\nu(E_x)\) is measurable, nor does it evaluate an integral of that function over \(X\). Those require further results. The useful point at this stage is that each individual section measure has an exact indicator-integral representation.
The Exact Formula for a Rectangle
Let \(A\in\mathcal M\) and \(B\in\mathcal N\). For the rectangle \(R=A\times B\), membership requires both \(x\in A\) and \(y\in B\). Thus the section is \(B\) when \(x\in A\), and is empty when \(x\notin A\). This gives an explicit formula for its indicator integral at every \(x\), without any finiteness assumption on \(\nu(B)\).
Proof. Fix \(x\). If \(x\in A\), then \((x,y)\in A\times B\) exactly when \(y\in B\), so \((A\times B)_x=B\). The indicator as a function of \(y\) is then \(\chi_B(y)\), whose integral is \(\nu(B)\). If \(x\notin A\), no pair \((x,y)\) belongs to \(A\times B\), so the section is empty and the indicator is identically zero. Its integral is zero. These two cases include all \(x\), and prove the formula. \(\square\)
When \(\mu(A)\) and \(\nu(B)\) are finite, the product measure of this rectangle is \((\mu\times\nu)(A\times B)=\mu(A)\nu(B)\), the defining rectangle property of the product measure. That fact is consistent with the section calculation, but the calculation above is only about a fixed section: it has not integrated the section values over \(X\).
Worked Example: A Rectangle with Finite Side Measures
Take \(X=Y=\mathbb R\) with Lebesgue measure \(m\), and let \(A=[-1,1]\), \(B=(2,5]\). These are Borel sets, and hence Lebesgue measurable. Their measures are \(m(A)=1-(-1)=2\) and \(m(B)=5-2=3\). For \(R=A\times B\), if \(-1\le x\le1\), then \(R_x=(2,5]\) and $$ \int_{\mathbb R}\chi_R(x,y)\,dm(y)=m((2,5])=5-2=3. $$ If \(x<-1\) or \(x>1\), then \(R_x=\varnothing\), and the same section integral is \(m(\varnothing)=0\). The endpoints \(x=-1\) and \(x=1\) are included because \(A\) is closed there.
Check: at \(x=0\), the section is \((2,5]\), of length \(5-2=3\); at \(x=2\), which is not in \([-1,1]\), the section is empty and the integral is \(0\). The product-measure rectangle formula gives \((m\times m)(R)=2\cdot3=6\), using the two side measures just computed. This is a separate product-measure calculation, not an interchange of integrals.
Sections of a General Measurable Set
For a set that is not a rectangle, its sections can vary with \(x\). The proposition still applies to every measurable set: at each fixed \(x\), the indicator integral is exactly the measure of that particular section. A section may be empty, have finite positive measure, or have infinite measure. Its value depends on the defining membership condition for the set.
Worked Example: Sections of an Open Triangle
On \(\mathbb R^2\), equipped in each coordinate with Lebesgue measure and with product measure \(m\times m\), consider the Borel set \(T=\{(x,y):0<y<x<1\}\). It is open, so it belongs to \(\mathcal B_{\mathbb R}\otimes\mathcal B_{\mathbb R}\), and therefore also to \(\mathcal L\otimes\mathcal L\). Fix \(x\). The conditions on \(y\) are \(0<y\) and \(y<x\), while \(x\) itself must satisfy \(x<1\). Consequently, $$ T_x= \begin{cases} (0,x),&0<x<1,\\ \varnothing,&x\le0\text{ or }x\ge1. \end{cases} $$ The section-indicator integral is \(m(T_x)=x\) when \(0<x<1\), and is zero otherwise. In particular, at \(x=\tfrac14\) the section is \((0,\tfrac14)\), so its measure and indicator integral are \(\tfrac14-0=\tfrac14\). At \(x=\tfrac34\) they are \(\tfrac34-0=\tfrac34\). At \(x=0\) and \(x=1\), the strict inequalities make the section empty.
Check: for \(x=\tfrac12\), the slice is \((0,\tfrac12)\), whose length is \(\tfrac12-0=\tfrac12\); the proposition gives \(\int_{\mathbb R}\chi_T(\tfrac12,y)\,dm(y)=\tfrac12\). This verifies the formula using the interval length directly. No integral over \(x\) has been computed here.
Worked Example: Every Section of the Diagonal Has Measure Zero
Let \(G=\{(x,y)\in\mathbb R^2:y=x\}\). This is a closed, hence Borel, subset of \(\mathbb R^2\). For every fixed \(x\in\mathbb R\), its section is \(G_x=\{x\}\). A singleton has Lebesgue measure zero, so $$ \int_{\mathbb R}\chi_G(x,y)\,dm(y)=m(\{x\})=0 $$ for every \(x\). The pointwise computation can also be checked directly: the indicator as a function of \(y\) is one at \(y=x\) and zero elsewhere, and changing a function at one point does not change its Lebesgue integral.
We can verify independently that \(G\) itself has product Lebesgue measure zero. Fix a positive integer \(N\). For any positive integer \(n\), divide \([-N,N]\) into \(n\) intervals of length \(2N/n\). The diagonal segment in \([-N,N]^2\) is covered by the \(n\) squares formed by pairing each interval with itself. Each square has area \((2N/n)^2\), so the total area of this cover is $$ n\left(\frac{2N}{n}\right)^2=\frac{4N^2}{n}. $$ As \(n\) increases this bound tends to zero; hence \(G\cap[-N,N]^2\) has outer measure zero and therefore product Lebesgue measure zero. Since \(G=\bigcup_{N=1}^\infty(G\cap[-N,N]^2)\), countable subadditivity gives \((m\times m)(G)=0\). Check: when \(N=2\) and \(n=8\), the interval length is \(4/8=1/2\), and the eight squares have total area \(8(1/2)^2=2\), matching \(4(2^2)/8=2\). The bound tends to zero as \(n\to\infty\), as required.
What the Section Calculation Does—and Does Not—Say
The indicator identity is pointwise in the fixed coordinate. It is not yet a theorem about the measure of an arbitrary set in the product space, because such a theorem would also need to justify measurability and integration of the section-measure function. Keep these tasks separate: first identify and measure each section; only a later product-integration result can justify passing from those values to an integral over \(X\).
A common error is to use the rectangle section \(B\) at every \(x\), forgetting that membership in \(A\) is also required. For the rectangle in the first example, claiming that the section at \(x=2\) is \((2,5]\) would give the incorrect calculation \(\int\chi_R(2,y)\,dm(y)=3\). But \(2\notin[-1,1]\), so \(R_2=\varnothing\), and the correct integral is \(0\). The correct rectangle formula is piecewise: the section is \(B\) only for \(x\in A\).
Another mistake is to turn the pointwise identity \(\int_Y\chi_E(x,y)\,d\nu(y)=\nu(E_x)\) directly into a formula for \((\mu\times\nu)(E)\). That step integrates over the other coordinate and requires an additional theorem; it does not follow just by rewriting the indicator. For rectangles the section calculation is explicit, and the product measure is separately determined by its rectangle property. For general sets, use the section identity as stated and do not interchange integrals without a theorem that permits it.
Check Your Understanding
For each question, first determine the section as a set, then use the indicator-integral identity.
- For \(R=[0,2]\times(3,7)\subset\mathbb R^2\), find \(R_x\) and \(\int_{\mathbb R}\chi_R(x,y)\,dm(y)\) for \(x=1\) and \(x=3\).
- For \(E=\{(x,y): -1<x<1,\ 0\le y\le x+1\}\), find \(E_x\) for \(-1<x<1\), and compute its measure at \(x=0\) and \(x=\tfrac12\).
- If \(E\in\mathcal M\otimes\mathcal N\), explain why \(\chi_E(x,y)=\chi_{E_x}(y)\) for a fixed \(x\), and state the resulting integral identity.
- For the graph \(G=\{(x,y):y=x\}\), identify \(G_x\) and its Lebesgue measure for each \(x\).
- Why does the section identity alone not justify integrating \(x\mapsto\nu(E_x)\) over \(X\) for an arbitrary measurable \(E\)?