From Simple Sections to General Functions
For a function \(f(x,y)\), fixing \(x\) gives a section \(y\mapsto f(x,y)\). If that section is nonnegative and measurable, its integral with respect to \(y\) is defined, possibly with value \(+\infty\). A key question is whether the resulting function of \(x\) is measurable. That property is needed before one can integrate the section-integral function or apply results about iterated integrals.
Let \((X,\mathcal{M},\mu)\) be any measure space, with no finiteness or sigma-finiteness assumption on \(\mu\), and let \((Y,\mathcal{N},\nu)\) be a sigma-finite measure space. Here sigma-finite means that \(Y\) is a countable union of sets of finite \(\nu\)-measure; it does not mean that \(\nu(Y)\) is finite. We use the product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\). No measure on the product space is needed for the measurability result below.
If \(f\) is \(\mathcal{M}\otimes\mathcal{N}\)-measurable, every section \(f_x\) is \(\mathcal{N}\)-measurable. Indeed, for fixed \(x\), the map \(y\mapsto(x,y)\) has measurable inverse images of rectangles \(A\times B\), since the inverse image is \(B\) if \(x\in A\) and is empty otherwise. The sets whose inverse images are measurable form a sigma-algebra, so the inverse image of every set in \(\mathcal{M}\otimes\mathcal{N}\) is measurable. Applying this to the inverse images of Borel sets under \(f\) gives the assertion for the sections.
Why Sigma-Finiteness Makes Simple Sections Measurable
The first step is a section-measure fact. It is where sigma-finiteness of \(\nu\), specifically, enters. The measure \(\mu\) on \(X\) plays no role in this fact.
Proof. Choose measurable sets \(Y_k\in\mathcal{N}\) with \(\nu(Y_k)<\infty\), \(Y_k\subseteq Y_{k+1}\), and \(\bigcup_{k=1}^{\infty}Y_k=Y\). Such an increasing cover is obtained from any countable finite-measure cover by taking successive finite unions. Fix \(k\). Let \(\mathcal{D}_k\) consist of the sets \(E\in\mathcal{M}\otimes\mathcal{N}\) for which \(x\mapsto\nu(E_x\cap Y_k)\) is measurable.
For a rectangle \(A\times B\), its section intersected with \(Y_k\) is \(B\cap Y_k\) when \(x\in A\), and is empty otherwise. Thus the function in question is \(\nu(B\cap Y_k)\chi_A(x)\), which is measurable and finite-valued. The rectangles form a pi-system: they contain \(X\times Y\) and are closed under finite intersections. The class \(\mathcal{D}_k\) is a lambda-system. It contains \(X\times Y\), because the corresponding function is the finite constant \(\nu(Y_k)\). If \(E\in\mathcal{D}_k\), then \(\nu(E_x^c\cap Y_k)=\nu(Y_k)-\nu(E_x\cap Y_k)\), so the relative complement also belongs to \(\mathcal{D}_k\). For pairwise disjoint \(E_j\in\mathcal{D}_k\), countable additivity gives \(\nu((\bigcup_j E_j)_x\cap Y_k)=\sum_j\nu((E_j)_x\cap Y_k)\); a pointwise sum of nonnegative measurable functions is measurable. The pi-lambda theorem therefore gives \(\mathcal{D}_k=\mathcal{M}\otimes\mathcal{N}\). Finally, for every \(E\in\mathcal{M}\otimes\mathcal{N}\), continuity from below of the measure \(\nu\) gives \(\nu(E_x)=\lim_{k\to\infty}\nu(E_x\cap Y_k)\). This is a pointwise limit of measurable functions, and hence is measurable. \(\square\)
Consequently, if \(\varphi=\sum_{j=1}^r a_j\chi_{E_j}\) is a nonnegative simple \(\mathcal{M}\otimes\mathcal{N}\)-measurable function, with disjoint measurable \(E_j\) and finite coefficients \(a_j\geq0\), then \(\int_Y\varphi(x,y)\,d\nu(y)=\sum_{j=1}^r a_j\nu((E_j)_x)\). Each summand is measurable by the lemma. A zero coefficient contributes zero, including on a section of infinite measure; no product \(0\cdot\infty\) is required.
Measurability for Every Nonnegative Function
Proof. We construct an increasing sequence of nonnegative simple measurable functions converging to \(f\). For \(n\geq1\), define \(\varphi_n(x,y)=2^{-n}\lfloor 2^n\min\{f(x,y),n\}\rfloor\). The minimum is interpreted as \(n\) when \(f(x,y)=+\infty\). Each \(\varphi_n\) is measurable and takes only the finitely many values \(0,2^{-n},\ldots,n\), so it is simple.
For completeness, these approximations increase. Write \(t=f(x,y)\). If \(t\leq n\), refining the dyadic mesh from \(2^{-n}\) to \(2^{-(n+1)}\) cannot decrease the rounded-down value. If \(t>n\), then \(\varphi_n(x,y)=n\), while \(\min\{t,n+1\}\geq n\), and \(n\) is itself a multiple of \(2^{-(n+1)}\); rounding down at that finer mesh gives a value at least \(n\). Thus \(\varphi_n\leq\varphi_{n+1}\). If \(t<\infty\), then for all \(n>t\) the truncation no longer changes \(t\), and \(0\leq t-\varphi_n(x,y)<2^{-n}\). If \(t=+\infty\), then \(\varphi_n(x,y)=n\). In both cases \(\varphi_n(x,y)\) increases to \(f(x,y)\).
Set \(F_n(x)=\int_Y\varphi_n(x,y)\,d\nu(y)\). The simple-function consequence of the lemma shows that every \(F_n\) is measurable. For each fixed \(x\), the functions \(y\mapsto\varphi_n(x,y)\) are nonnegative, measurable, increase pointwise to \(f_x\), and are integrated on the measure space \((Y,\mathcal{N},\nu)\). The Monotone Convergence Theorem, which does not require finite or sigma-finite measure, therefore gives \(F(x)=\lim_{n\to\infty}F_n(x)\). A pointwise limit of measurable extended nonnegative functions is measurable. This proves the theorem. The limit and integral have been related only on each fixed section, by Monotone Convergence; no interchange of integrals is asserted here. \(\square\)
Worked Example: A Simple Function on a Counting Measure Space
Let \(X=\{a,b,c\}\) with its power-set sigma-algebra and counting measure \(\mu\), which is finite. Let \(Y=\mathbb{N}=\{1,2,\ldots\}\) with its power-set sigma-algebra and counting measure \(\nu\), which is sigma-finite but not finite. Define \(f=2\chi_{\{a\}\times\{1\}}+\chi_{\{a\}\times\{2\}}+3\chi_{\{b\}\times\{1\}}\). This is a nonnegative simple product-measurable function.
For \(x=a\), the section has value \(2\) at \(1\), value \(1\) at \(2\), and value \(0\) at every other positive integer. Its counting-measure integral is the sum of its values, so \(F(a)=2+1=3\). For \(x=b\), the only nonzero value is \(f(b,1)=3\), so \(F(b)=3\). For \(x=c\), every value is zero, so \(F(c)=0\). Thus \(F\) is the function with values \(3,3,0\) at \(a,b,c\), respectively, and is measurable on the power-set sigma-algebra. Check: summing the displayed simple-function coefficients over the \(Y\)-sections gives \(2+1=3\) at \(a\), \(3\) at \(b\), and no nonzero terms at \(c\), in agreement with the section calculations.
Worked Example: Sections of a Triangle
Take \(X=\mathbb{R}\) with its Borel sigma-algebra and Lebesgue measure \(m\), and \(Y=[0,\infty)\) with its Borel sigma-algebra and Lebesgue measure. Both measures are sigma-finite. Let \(f(x,y)=\chi_E(x,y)\), where \(E=\{(x,y):0\leq y\leq x\}\). The set \(E\) is closed, hence Borel in \(\mathbb{R}\times[0,\infty)\), so \(f\) is product-measurable.
If \(x<0\), there are no \(y\in[0,\infty)\) with \(y\leq x\); the section is empty and \(F(x)=0\). If \(x=0\), the section is the singleton \(\{0\}\), which has Lebesgue measure zero, so \(F(0)=0\). If \(x>0\), the section is the closed interval \([0,x]\), whose Lebesgue measure is \(x\). Therefore \(F(x)=\max\{x,0\}\), a Borel-measurable function. Check: at \(x=-2,0,3\), the section measures are \(0,0,3\), respectively, exactly the values of \(\max\{x,0\}\). The endpoint convention is \(0\leq y\leq x\); changing either endpoint would not change these Lebesgue measures.
Worked Example: Integrating an Exponential Section
Let \(X=Y=(0,\infty)\), each with its Borel sigma-algebra and Lebesgue measure, and set \(f(x,y)=e^{-xy}\). This function is continuous, hence product-measurable, and is nonnegative. For any fixed \(x>0\), substitute \(u=xy\), so \(dy=du/x\). Then \(\displaystyle F(x)=\int_0^\infty e^{-xy}\,dy=\frac1x\int_0^\infty e^{-u}\,du=\frac1x\bigl[-e^{-u}\bigr]_{0}^{\infty}=\frac1x\). This is finite for each \(x>0\) and is a continuous, hence Borel-measurable, function of \(x\). Check: at \(x=2\), substitution gives \(\frac12\int_0^\infty e^{-u}\,du=\frac12(1)=\frac12\), matching \(1/x=1/2\). The general theorem also gives measurability without this explicit evaluation.
The Sigma-Finiteness Pitfall
A common mistake is to claim that section-integral measurability holds for every measure \(\nu\), then use the complement identity \(\nu(E_x^c)=\nu(Y)-\nu(E_x)\) in a proof. When \(\nu(Y)=\infty\) and \(\nu(E_x)=\infty\), this calculation becomes \(\infty-\infty\), which is undefined. Restricting first to finite-measure sets \(Y_k\) makes the valid calculation \(\nu(Y_k)-\nu(E_x\cap Y_k)\); sigma-finiteness then lets those restrictions exhaust \(Y\).
The hypothesis cannot simply be dropped. A standard descriptive-set-theoretic fact is that for \(X=Y=\mathbb{N}^{\mathbb{N}}\) with their Borel sigma-algebras, there is a Borel set \(E\subseteq X\times Y\) whose projection \(A=\{x:E_x\ne\varnothing\}\) is analytic but not Borel. Give \(Y\) the counting measure on its Borel sets: \(\nu(B)\) is the number of points of \(B\), or \(+\infty\) if \(B\) is infinite. This measure is not sigma-finite, since its finite-measure sets are finite and a countable union of finite sets cannot cover the uncountable space \(Y\). For \(f=\chi_E\), the section integral is \(F(x)=\nu(E_x)\), and \(\{x:F(x)>0\}=A\). Since \(A\) is not Borel, \(F\) is not Borel-measurable. Thus sigma-finiteness of the integrating measure \(\nu\) is a genuine condition for the stated theorem; no corresponding sigma-finiteness condition on \(\mu\) is needed.
This result establishes measurability of the section-integral function. It does not, by itself, equate its integral over \(X\) with an integral over the product space, nor does it assert an interchange of integrals. Such identities require additional theorems and hypotheses.
Check Your Understanding
For each question, identify the measure space and check the role of sigma-finiteness before applying the result.
- In the theorem, which of the two measures must be sigma-finite, and where is that assumption used in the proof?
- For \(f(x,y)=\chi_{\{(x,y):0\leq y\leq x\}}\) with \(y\in[0,\infty)\), compute the section integral at \(x=-1\), \(x=0\), and \(x=4\).
- Explain why \(2^{-n}\lfloor2^n\min\{t,n\}\rfloor\) increases to \(t\) for finite \(t\geq0\), and increases without bound when \(t=+\infty\).
- In the simple-function example on \(\mathbb{N}\) with counting measure, why is the integral of the section at \(a\) equal to \(3\)?
- Why is the complement calculation \(\nu(Y)-\nu(E_x)\) not valid when both terms are infinite, and how does a finite-measure exhaustion repair it?