Measurability of integration against a kernel

First suppose K finite and write $m(s)=K(s,T)<\mathrm{\infty }$. For a product-measurable set E, its section $E_s$ is measurable, by applying the section theorem to its indicator. Let $\mathcal{D}$ consist of those E for which $s\mapsto K(s,E_s)$ is measurable. It contains every rectangle $A\times B$, whose evaluation is $1_A(s)K(s,B)$, and it contains $S\times T$. If E belongs to this class, then $(E^c{)}_s=T\setminus E_s$ and $K(s,(E^c{)}_s)=m(s)-K(s,E_s)$; both terms are finite measurable real functions, so the difference is measurable. If E_j are disjoint class members, their sections are disjoint and $K(s,({\bigcup }_j{E}_j{)}_s)={\sum }_{j}K(s,(E_j{)}_s)$ is an increasing limit of measurable finite sums. Thus this class is a lambda-system. Rectangles form a pi-system, so Dynkin's theorem gives every product-measurable E in the class. The measurable-closure proposition can be used on S equipped with its zero measure; its conclusions concern only Sigma and do not require a preexisting source probability.

For a nonnegative simple product-measurable function $g={\sum }_{j=1}^r{a}_j{1}_{E_j}$ with disjoint E_j and finite nonnegative coefficients, its section integral is ${\sum }_j{a}_{j}K(s,(E_j{)}_s)$ and is measurable by step 1.1. Choose the prescribed increasing simple approximation $g_n\uparrow f$ on the product. For every s the section theorem and monotone convergence give $I_f(s)={\lim }_n{I}_{g_n}(s)$, so the increasing-limit closure proves measurability. No uniform bound in s was used; only the individual finite masses entered the complement calculation.

Now let $(T_n)$ be the specified common exhaustion. Define $K_n(s,A)=K(s,A\cap T_n)$. For each s this is the restriction of a measure, its evaluations are measurable by the kernel hypothesis, and $K_n(s,T)=K(s,T_n)<\mathrm{\infty }$. Apply step 2.1 to K_n. Sectionwise integration against the restriction equals integration of $f(s,\cdot )1_{T_n}$ against K(s,·): this holds for indicators by definition, for simple functions by finite sums, and for nonnegative functions by monotone convergence. Since $T_n\uparrow T$, $I_f(s)={\lim }_n\int f(s,t)K_n(s,dt)$; another increasing-limit argument proves the result. If T is empty all these integrals are zero; if S is empty the assertion is vacuous.

For real f its positive and negative parts and absolute value are product-measurable. The proved result makes $I_{f^{+}},I_{f^{-}},I_{\mathrm{\mid }f\mathrm{\mid }}$ measurable. Therefore $D={\bigcup }_{n\ge 1}\{I_{\mathrm{\mid }f\mathrm{\mid }}<n\}$ is measurable. On D both part integrals are finite since each is at most $I_{\mathrm{\mid }f\mathrm{\mid }}$. Define $J_{+}(s)=I_{f^{+}}(s)$ on D and zero otherwise, and define J_- similarly. Nonnegative measurable restriction makes both J_± measurable; they are finite everywhere. Their real difference is the requested signed integral on D and zero elsewhere. This never subtracts two infinities. For a zero kernel all integrals vanish and D=S, even if f is unbounded; for f=0 the same holds for every permitted kernel. All approximations and the exhaustion are specified; no AC is used.