From Section Sets to Section Integrals
For a function on a product space, fixing the outer variable leaves a function of the inner variable. Integrating that function gives a number depending on the variable we fixed. For a nonnegative simple function, this dependence can be described by a finite sum. The key point to justify is that the sets measured in that sum vary measurably with the outer variable.
Throughout, let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces, and use the product measurable space \((X\times Y,\mathcal{M}\otimes\mathcal{N})\) and product measure \(\mu\times\nu\). We do not require either measure to be finite or either space to be complete. For a set \(E\in\mathcal{M}\otimes\mathcal{N}\) and \(x\in X\), its section is \(E_x=\{y\in Y:(x,y)\in E\}\). For a function \(f\) on \(X\times Y\), write \(f_x(y)=f(x,y)\). Sections of product-measurable sets are \(\mathcal{N}\)-measurable, as established in the earlier material on sections.
The disjoint-set representation makes the calculation direct. For each \(x\), the section is \(f_x=\sum_{j=1}^k a_j\chi_{(E_j)_x}\). The definition of the integral of a nonnegative simple function therefore gives
This equality holds for every \(x\in X\), including when one or more section measures are infinite. Since the sets \(E_j\) are disjoint, their sections are disjoint for each fixed \(x\). That condition ensures the displayed representation computes \(f_x\) without counting any point twice.
Why the Section Integral Is Measurable
The required set-level fact is that, for product-measurable \(E\), the function \(x\mapsto\nu(E_x)\) is \(\mathcal{M}\)-measurable. Here is a proof that also identifies why sigma-finiteness of \(\nu\) is useful. A pi-system is a family of sets closed under finite intersections. A lambda-system contains the whole space, is closed under complements, and is closed under countable disjoint unions. We use the pi-lambda theorem, in the form that a lambda-system containing a pi-system contains the sigma-algebra generated by that pi-system.
Proof. First suppose \(\nu(Y)<\infty\), and let \(\mathcal{D}\) be the class of sets \(E\in\mathcal{M}\otimes\mathcal{N}\) for which \(x\mapsto\nu(E_x)\) is \(\mathcal{M}\)-measurable. The whole space belongs to \(\mathcal{D}\), since its section measure is the constant \(\nu(Y)\). If \(E\in\mathcal{D}\), then \((E^c)_x=Y\setminus E_x\), so \(\nu((E^c)_x)=\nu(Y)-\nu(E_x)\); this is measurable because \(\nu(Y)\) is finite. If \(E_1,E_2,\ldots\) are pairwise disjoint members of \(\mathcal{D}\), their sections are pairwise disjoint and \(\nu((\bigcup_n E_n)_x)=\sum_n\nu((E_n)_x)\). The partial sums are measurable and increase to this sum, so the sum is measurable. Thus \(\mathcal{D}\) is a lambda-system.
Measurable rectangles \(A\times B\), with \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\), form a pi-system containing \(X\times Y\). For such a rectangle, \((A\times B)_x=B\) if \(x\in A\), and is empty otherwise; consequently \(x\mapsto\nu((A\times B)_x)=\nu(B)\chi_A(x)\) is measurable. The pi-lambda theorem now gives \(\mathcal{D}=\mathcal{M}\otimes\mathcal{N}\).
For general sigma-finite \(\nu\), choose measurable sets \(Y_n\) with \(\nu(Y_n)<\infty\), \(Y_n\subseteq Y_{n+1}\), and \(\bigcup_nY_n=Y\). For a product-measurable \(E\), each \(E\cap(X\times Y_n)\) is product-measurable, so the finite-measure case shows that \(x\mapsto\nu(E_x\cap Y_n)\) is measurable. For each \(x\), continuity from below of measures gives \(\nu(E_x)=\lim_n\nu(E_x\cap Y_n)\). A pointwise limit of measurable functions is measurable, including when the limit is infinite. This proves the set-level fact. \(\square\)
Proof. Fix \(x\in X\). Restricting the finite disjoint representation of \(f\) to the section gives \(f_x=\sum_{j=1}^k a_j\chi_{(E_j)_x}\), again with pairwise disjoint sets. The definition and finite additivity of the integral of a nonnegative simple function give the stated formula, with infinite values allowed. By the set-level fact just proved, each function \(x\mapsto\nu((E_j)_x)\) is \(\mathcal{M}\)-measurable. Multiplication by the finite constant \(a_j\ge0\) and addition of finitely many nonnegative measurable functions preserve measurability. Therefore \(G\) is measurable on all of \(X\). \(\square\)
Worked Section Calculations
Worked Example: Sections on Finite Counting Spaces
Let \(X=\{1,2,3\}\) have counting measure \(\mu\), and let \(Y=\{a,b,c\}\) have measure \(\nu\) with \(\nu(\{a\})=1\), \(\nu(\{b\})=2\), and \(\nu(\{c\})=1\). Define \(f\) by the following values, with rows indexed by \(x\) and columns by \(y\):
| \(f(x,y)\) | \(a\) | \(b\) | \(c\) |
|---|---|---|---|
| \(x=1\) | 2 | 0 | 1 |
| \(x=2\) | 0 | 3 | 1 |
| \(x=3\) | 1 | 1 | 0 |
Every subset is measurable, so \(f\) is a nonnegative simple function on this finite product. The integral against \(\nu\) is the weighted sum over the three points of \(Y\). For \(x=1\), it is \(2\cdot1+0\cdot2+1\cdot1=3\). For \(x=2\), it is \(0\cdot1+3\cdot2+1\cdot1=7\). For \(x=3\), it is \(1\cdot1+1\cdot2+0\cdot1=3\). Thus \(G(1)=3\), \(G(2)=7\), and \(G(3)=3\). Check: the three weights sum to \(1+2+1=4\), and direct row-by-row weighted sums give \(3,7,3\), respectively. Since every subset of \(X\) is measurable, this function \(G\) is measurable.
Worked Example: Overlapping Outer Supports
On \(\mathbb{R}\times\mathbb{R}\) with Lebesgue measure \(m\) on each factor, let \(f=2\chi_{[0,2)\times[1,3]}+5\chi_{[1,3]\times[4,5)}\). The two product sets are disjoint because their \(y\)-intervals are disjoint. The first has section \([1,3]\) for \(x\in[0,2)\), of length \(3-1=2\); the second has section \([4,5)\) for \(x\in[1,3]\), of length \(5-4=1\). The section integral is therefore
Indeed, on \([1,2)\) both terms contribute, giving \(2\cdot2+5\cdot1=9\); on \([0,1)\) only the first contributes, giving \(2\cdot2=4\); on \([2,3]\) only the second contributes, giving \(5\cdot1=5\). At \(x=2\), the first support excludes \(x\), while the second includes it, so the value there is \(5\). At \(x=3\), the second support includes its endpoint, and its section still has length \(1\). Outside \([0,3]\) both sections are empty and the integral is \(0\). Check: these four cases are intervals and their complements, hence Borel sets; the displayed piecewise-constant \(G\) is measurable.
Worked Example: A Triangular Section
On \(\mathbb{R}^2\) with Lebesgue measure in both coordinates, set \(E=\{(x,y):0\le y\le x\le1\}\), \(R=[0,1]\times[2,3]\), and \(f=3\chi_E+2\chi_R\). Both sets are Borel, and they are disjoint because points of \(E\) have \(y\le1\), whereas points of \(R\) have \(y\ge2\). For \(x\in[0,1]\), the section \(E_x=[0,x]\) has length \(x\), and \(R_x=[2,3]\) has length \(3-2=1\). Thus \(G(x)=3x+2\) on \([0,1]\). For \(x\notin[0,1]\), both sections are empty, so \(G(x)=0\).
At \(x=0\), the section \(E_0=\{0\}\) has Lebesgue measure zero, giving \(G(0)=3\cdot0+2\cdot1=2\). At \(x=\tfrac12\), the two contributions are \(3\cdot\tfrac12=\tfrac32\) and \(2\cdot1=2\), so \(G(\tfrac12)=\tfrac72\). At \(x=1\), they are \(3\cdot1=3\) and \(2\cdot1=2\), so \(G(1)=5\). Check: the formula \(3x+2\) gives \(2,\tfrac72,5\) at those three points. The function is measurable because it is linear on the measurable interval \([0,1]\) and zero outside it.
A Common Counting Error
A finite sum of indicators need not be a disjoint-set representation. For example, if \(A=B=[0,1]\times[0,1]\), then \(\chi_A+\chi_B=2\chi_A\), not \(\chi_{A\cup B}\). For each \(x\in[0,1]\), the section of each set is \([0,1]\), whose measure is \(1\). The section integral of the sum is therefore \(1+1=2\), not \(1\). Outside \([0,1]\), it is \(0\).
The correction is to respect the function’s actual values on overlaps. In this case the overlap has value \(2\), so the disjoint representation is \(2\chi_A\), whose section integral is \(2\cdot1=2\) on \([0,1]\). More generally, to use a disjoint representation, partition the union of the sets according to which original sets contain each point, then attach the corresponding sum of coefficients to each piece. The formula for section integrals applies to those disjoint pieces, with no accidental omission or double counting.
The result here concerns finite-valued nonnegative simple functions. It establishes measurability by reducing their section integrals to finitely many section-measure functions. Extending that measurability argument to general nonnegative measurable functions requires an additional approximation step and is a separate result.
Check Your Understanding
In each question, specify the section at the stated outer point before computing its integral.
- On finite counting spaces, suppose \(\nu(\{a\})=2\) and \(\nu(\{b\})=1\), while \(f_x(a)=3\) and \(f_x(b)=4\). Compute \(\int_Y f_x\,d\nu\).
- For \(f=4\chi_{[0,1]\times[0,2]}+\chi_{[1,2]\times[3,4]}\) on \(\mathbb{R}^2\), compute the section integral at \(x=\tfrac12\), \(x=1\), and \(x=2\), paying attention to endpoint inclusion.
- Explain why the proof of measurability for \(x\mapsto\nu(E_x)\) first treats finite \(\nu\), and how sigma-finiteness extends the result.
- If \(E_1,E_2\) are disjoint product-measurable sets, show from the simple-function integral formula that the section integral of \(a\chi_{E_1}+b\chi_{E_2}\), for \(a,b\ge0\), is \(a\nu((E_1)_x)+b\nu((E_2)_x)\).
- For \(A=B=[0,1]\times[0,1]\), compute the section integral of \(\chi_A+\chi_B\) at \(x=\tfrac12\). Why is the integral not the section length of \(A\cup B\)?