Measurability of Section Integrals

Why Measurability of Section Integrals Matters

Suppose a function \(f(x,y)\) is measurable on a product space. For each fixed \(x\), we can try to integrate over the \(y\)-coordinate and obtain a new function of \(x\). Before using that function in another integral or probability calculation, we need to know that it is measurable.

This is not automatic from the notation alone. The key step is first to understand sections of measurable sets: for a product-measurable set \(E\), how does the measure of its \(y\)-section vary with \(x\)? Sigma-finiteness of the measure in the integrated coordinate makes that dependence measurable.

Standing assumptions: Let \((X,\mathcal A)\) and \((Y,\mathcal B)\) be measurable spaces, and let \(\nu\) be a sigma-finite measure on \((Y,\mathcal B)\). For \(E\in\mathcal A\otimes\mathcal B\), define the section \(E_x=\{y\in Y:(x,y)\in E\}\). For a product-measurable function \(f\), write \(f_x(y)=f(x,y)\).

Measurable Sections and Their Measures

The section \(E_x\) is in \(\mathcal B\) for every \(x\) when \(E\) belongs to the product sigma-algebra. This is the measurable-sections result. The further claim that \(x\mapsto\nu(E_x)\) is measurable requires an argument.

Theorem: If \(\nu\) is sigma-finite and \(E\in\mathcal A\otimes\mathcal B\), then the function \(x\mapsto\nu(E_x)\), with values in \([0,\infty]\), is \(\mathcal A\)-measurable.

Proof. First suppose \(\nu(Y)<\infty\). Let \(\mathcal C\) be the class of all \(E\in\mathcal A\otimes\mathcal B\) for which \(x\mapsto\nu(E_x)\) is measurable. If \(E=A\times B\) is a measurable rectangle, then \(\nu(E_x)=\nu(B)\mathbf 1_A(x)\), so \(E\in\mathcal C\). Rectangles form a pi-system that generates \(\mathcal A\otimes\mathcal B\).

The class \(\mathcal C\) is a Dynkin system. It contains \(X\times Y\), since the corresponding function is the constant \(\nu(Y)\). If \(E\in\mathcal C\), then \(\nu((E^c)_x)=\nu(Y)-\nu(E_x)\), which is measurable because \(\nu(Y)\) is finite. If \(E_1,E_2,\ldots\) are pairwise disjoint members of \(\mathcal C\), then \(\nu((\bigcup_{j=1}^{\infty}E_j)_x)=\sum_{j=1}^{\infty}\nu((E_j)_x)\), a pointwise limit of measurable partial sums. Thus the pi-lambda theorem gives \(\mathcal C=\mathcal A\otimes\mathcal B\).

Now let \(\nu\) be sigma-finite. Choose a countable cover of \(Y\) by measurable sets of finite measure, and make it into a disjoint measurable partition \(Y_1,Y_2,\ldots\), each member having finite measure. For any \(E\in\mathcal A\otimes\mathcal B\), the finite-measure case shows that \(x\mapsto\nu(E_x\cap Y_n)\) is measurable for every \(n\). Since the pieces are disjoint, \(\nu(E_x)=\sum_{n=1}^{\infty}\nu(E_x\cap Y_n)\). The right side is a limit of measurable partial sums, so it is measurable, including when its value is infinity. \(\square\)

The finite-measure argument uses complements by subtracting from \(\nu(Y)\). The partition argument lets us use that reasoning locally when the full measure of \(Y\) is infinite. This is where sigma-finiteness enters.

Worked Example: Sections of a Subgraph

Take \(X=Y=[0,1]\) with their Borel sigma-algebras and Lebesgue measure \(\nu\) on \(Y\). Let \(E=\{(x,y):0\leq y\leq x\leq1\}\). This is a Borel subset of the square, and for each \(x\in[0,1]\), its section is \(E_x=[0,x]\). Therefore \(\nu(E_x)=x\). The section-measure theorem ensures measurability of this function; the direct calculation also identifies it as the identity function on \([0,1]\).

At the endpoints, \(\nu(E_0)=\nu(\{0\})=0\) and \(\nu(E_1)=\nu([0,1])=1\), agreeing with the formula \(x\).

Nonnegative Section Integrals

For a nonnegative product-measurable function, the section integral is defined for every \(x\), though it may be infinite. The passage from sets to functions follows by approximating a nonnegative measurable function from below by measurable simple functions.

Theorem: If \(f:X\times Y\to[0,\infty]\) is \(\mathcal A\otimes\mathcal B\)-measurable and \(\nu\) is sigma-finite, then \(F(x)=\int_Y f(x,y)\,d\nu(y)\) is an \(\mathcal A\)-measurable function from \(X\) to \([0,\infty]\).

Proof. For an indicator function \(f=\mathbf 1_E\), the section integral is \(\int_Y\mathbf 1_E(x,y)\,d\nu(y)=\nu(E_x)\), which is measurable by the preceding theorem.

Now let \(s=\sum_{j=1}^m a_j\mathbf 1_{E_j}\) be a nonnegative measurable simple function, where each \(a_j\geq0\) and each \(E_j\) is product-measurable. Linearity of the nonnegative integral gives \(\int_Y s(x,y)\,d\nu(y)=\sum_{j=1}^m a_j\nu((E_j)_x)\). Each term is measurable, so the finite sum is measurable.

Finally, choose nonnegative measurable simple functions \(s_n\) increasing pointwise to \(f\). For every fixed \(x\), the monotone convergence property of the nonnegative integral gives \(\int_Y f(x,y)\,d\nu(y)=\lim_{n\to\infty}\int_Y s_n(x,y)\,d\nu(y)\). Each function in this limit is measurable, and a pointwise limit of measurable extended nonnegative functions is measurable. Hence \(F\) is measurable. \(\square\)

This theorem asserts measurability, not finiteness. An infinite value at some \(x\) is permitted and does not prevent the section-integral function from being measurable.

Worked Example: An Infinite Value at One Section

Let \(X=Y=[0,\infty)\) with Borel sigma-algebras and Lebesgue measure on \(Y\), and set \(f(x,y)=e^{-xy}\). This is a nonnegative continuous, hence product-measurable, function. At \(x=0\), \(f(0,y)=1\), so \(\int_0^\infty f(0,y)\,dy=\infty\). For \(x>0\), direct integration gives \(\int_0^\infty e^{-xy}\,dy=[-e^{-xy}/x]_{y=0}^{y=\infty}=1/x\).

Thus the section integral is \(+\infty\) at zero and \(1/x\) for \(x>0\). Its extended-valued measurability is consistent with the theorem, even though it is not finite everywhere.

Signed Functions and Absolute Integrability of Sections

For a real-valued measurable function, write \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\). Both are nonnegative and measurable, and \(f=f^+-f^-\) while \(|f|=f^++f^-\). Each of their section integrals is measurable by the nonnegative result.

Theorem: Let \(f:X\times Y\to\mathbb R\) be product-measurable, with \(\nu\) sigma-finite. The set \(D=\{x:\int_Y |f(x,y)|\,d\nu(y)<\infty\}\) is measurable. On \(D\), the section integral \(H(x)=\int_Y f(x,y)\,d\nu(y)\) is finite and measurable. Defining \(H(x)=0\) for \(x\notin D\) gives a measurable real-valued function on all of \(X\).

Proof. By the nonnegative theorem, the functions \(A(x)=\int_Y f^+(x,y)\,d\nu(y)\) and \(B(x)=\int_Y f^-(x,y)\,d\nu(y)\) are measurable and extended nonnegative. Since \(|f|=f^++f^-\), its section integral is \(A+B\), so \(D=\{x:A(x)+B(x)<\infty\}\) is measurable.

For \(x\in D\), both \(A(x)\) and \(B(x)\) are finite. Consequently the signed integral exists absolutely and equals \(A(x)-B(x)\). The functions \(A\) and \(B\), restricted to \(D\), are finite measurable functions, so their difference is measurable on \(D\). Extending that difference by zero on the measurable set \(D^c\) preserves measurability: for any open set \(U\subseteq\mathbb R\), its preimage is the union of the preimage within \(D\) and, if \(0\in U\), the set \(D^c\). Both parts are measurable. This proves the claim. \(\square\)

The condition defining \(D\) is absolute integrability of the section. It rules out the undefined expression \(+\infty-\infty\), which can occur if one tries to subtract the two nonnegative section integrals without checking finiteness.

Worked Example: A Signed Section Integral

Let \(X=Y=[0,1]\) with Lebesgue measure and define \(f(x,y)=x-y\). This function is continuous and bounded, so every section is absolutely integrable. For each \(x\in[0,1]\), \(\int_0^1 |x-y|\,dy\leq\int_0^1 1\,dy=1\), which verifies that \(x\in D\).

Splitting the signed integral into its two terms gives \(\int_0^1(x-y)\,dy=x\int_0^1 1\,dy-\int_0^1 y\,dy=x-\tfrac12\). The resulting function is measurable (in fact, continuous). At \(x=0\), the value is \(-\tfrac12\); at \(x=1\), it is \(\tfrac12\), as the formula predicts.

Worked Example: A Section Integral with a Measurable Support

Let \(X=Y=\mathbb R\) with their Borel sigma-algebras and Lebesgue measure on \(Y\), and set \(f(x,y)=\mathbf 1_{\{(x,y):\,x<y<x+1\}}\). The defining region is open, so \(f\) is product-measurable and nonnegative. For fixed \(x\), the section is \((x,x+1)\), whose Lebesgue measure is \(1\). Hence \(\int_{\mathbb R} f(x,y)\,dy=1\) for every \(x\), a measurable function.

This example also shows that the theorem applies when the support moves with \(x\): measurability does not require the sections to be identical or nested.

What This Result Does—and Does Not—Say

Measurability of \(x\mapsto\int f(x,y)\,d\nu(y)\) is a prerequisite for integrating that function in \(x\), but it does not by itself justify exchanging the order of integration. Order-switching theorems require further hypotheses and establish additional conclusions.

For signed functions, the result above deliberately describes the measurable set \(D\) where sections are absolutely integrable. If \(f\) is integrable on a product measure space, a theorem on nonnegative iterated integrals will later provide the usual almost-everywhere finiteness statement for its sections. That finiteness conclusion is separate from the measurability argument proved here.

Key takeaway: Sigma-finiteness of \(\nu\) gives measurability of section measures. Approximation by simple functions then gives measurability of nonnegative section integrals, and positive and negative parts give the signed result wherever the sections are absolutely integrable.

Check Your Understanding

Use the definitions and proofs above to answer these questions.

Where does the proof of measurability of \(x\mapsto\nu(E_x)\) use finiteness of the measure of a set?
Why is the section integral of a nonnegative function allowed to take the value \(+\infty\)?
For a real-valued measurable function \(f\), express the set of absolutely integrable sections using the section integrals of \(f^+\) and \(f^-\).
Why can the expression \(\int f^+(x,y)\,d\nu(y)-\int f^-(x,y)\,d\nu(y)\) fail to define a signed integral for some \(x\)?
Does measurability of a section-integral function alone justify exchanging the order of integration? Explain briefly.