Proof of Integration Against a Probability Kernel: Measurable Sections, the Composite Measure on the Product and the Iterated Integral

Each result cited is universally quantified over the data in its own statement.

Throughout, $Y\times Z$ carries the $\sigma$-algebra $\mathcal{Y}\otimes\mathcal{Z}$, measurability of a real-valued function is with respect to the Borel $\sigma$-algebra of the real line as in Measure Spaces and the Lebesgue Integral: Standing Notation §measurable, and integrals are those of Measure Spaces and the Lebesgue Integral: Standing Notation §integral. A nonnegative measurable real-valued function is also a measurable map into $[0,\infty]$ in the sense of Lebesgue Integral of a Nonnegative Measurable Function, by Measure Spaces and the Lebesgue Integral: Standing Notation §measurable. By Probability Kernels Between Measurable Spaces §kernel, each $\kappa_{y}$ is a probability measure on $(Z,\mathcal{Z})$ and each $\kappa(\cdot,B)$, $B\in\mathcal{Z}$, is measurable with respect to $\mathcal{Y}$.

Step 0: two facts about a finite measure. Let $(X,\mathcal{F},\lambda_{0})$ be a measure space with $\lambda_{0}(X)$ a real number.

(P1) Let $h:X\to\mathbb{R}$ be measurable and let $c\ge0$ be real with $|h(x)|\le c$ for every $x\in X$. Then $h$ is integrable and $\bigl|\int_{X}h\,d\lambda_{0}\bigr|\le c\,\lambda_{0}(X)$. Indeed, $|h|$ is measurable by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and $|h|\le c\mathbf{1}_{X}$, so Linearity and Monotonicity of the Lebesgue Integral §nonnegative (monotonicity and the scalar rule) and The Integral of an Indicator Function is the Measure of the Set give $\int_{X}|h|\,d\lambda_{0}\le c\int_{X}\mathbf{1}_{X}\,d\lambda_{0}=c\,\lambda_{0}(X)<\infty$; hence $h$ is integrable by the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §integral, and $\bigl|\int_{X}h\,d\lambda_{0}\bigr|\le\int_{X}|h|\,d\lambda_{0}$ by Linearity and Monotonicity of the Lebesgue Integral §integrable; here the Lebesgue integral of the nonnegative integrable function $|h|$ agrees with its integral as a nonnegative measurable function, bounded above, by (P2) below (whose proof does not use (P1)).

(P2) Let $h:X\to\mathbb{R}$ be measurable and integrable with $h\ge0$. Then its Lebesgue integral equals its integral as a nonnegative measurable function. Indeed, in the notation of Integrable Function and the Lebesgue Integral, $h^{+}=h$ and $h^{-}$ is the zero function $0\cdot\mathbf{1}_{X}$, whose integral is $0\cdot\lambda_{0}(X)=0$ by Linearity and Monotonicity of the Lebesgue Integral §nonnegative and The Integral of an Indicator Function is the Measure of the Set.

Step 1: clause 1 (Sections). For $E\in\mathcal{Y}\otimes\mathcal{Z}$ write $h_{E}(y)=\kappa(y,E_{y})$ for $y\in Y$; this is defined since $E_{y}\in\mathcal{Z}$ by claim 1 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable. Since $\kappa_{y}$ is a measure with $\kappa_{y}(Z)=1$ and $E_{y}\subseteq Z$, Basic Properties of a Measure §monotone gives $0\le h_{E}(y)\le1$.

Let $\mathcal{P}$ be the family of measurable rectangles $A\times B$ with $A\in\mathcal{Y}$ and $B\in\mathcal{Z}$, and let $\mathcal{D}$ be the family of those $E\in\mathcal{Y}\otimes\mathcal{Z}$ for which $h_{E}$ is measurable with respect to $\mathcal{Y}$. We apply Dynkin's Pi-Lambda Theorem with the $\pi$-system $\mathcal{P}$ and the $\lambda$-system $\mathcal{D}$, both families of subsets of $Y\times Z$.

$\mathcal{P}$ is a $\pi$-system: it contains $Y\times Z$, and $(A\times B)\cap(A'\times B')=(A\cap A')\times(B\cap B')$ with $A\cap A'\in\mathcal{Y}$ and $B\cap B'\in\mathcal{Z}$ by Sigma-Algebra and Measurable Space.

$\mathcal{P}\subseteq\mathcal{D}$: by Product Sigma-Algebra and Generated Sigma-Algebra, $\mathcal{P}\subseteq\mathcal{Y}\otimes\mathcal{Z}$. For $A\in\mathcal{Y}$, $B\in\mathcal{Z}$ and $y\in Y$, the section $(A\times B)_{y}$ is $B$ if $y\in A$ and $\varnothing$ if $y\notin A$; since $\kappa(y,\varnothing)=0$,

$$h_{A\times B}(y)=\mathbf{1}_{A}(y)\,\kappa(y,B)\qquad(y\in Y),\tag{1}$$

which is measurable by claims 1 and 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and the measurability of $\kappa(\cdot,B)$.

$\mathcal{D}$ is a $\lambda$-system. ($\lambda$1) $Y\times Z\in\mathcal{D}$, because $h_{Y\times Z}(y)=\kappa(y,Z)=1$ for every $y$, and constants are measurable by claim 1 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. ($\lambda$2) Let $E,F\in\mathcal{D}$ with $E\subseteq F$. Then $F\setminus E=F\cap((Y\times Z)\setminus E)\in\mathcal{Y}\otimes\mathcal{Z}$, and for each $y$ we have $(F\setminus E)_{y}=F_{y}\setminus E_{y}$ with $E_{y}\subseteq F_{y}$ and $\kappa_{y}(F_{y})\le1<\infty$, so Basic Properties of a Measure §differences gives $h_{F\setminus E}(y)=h_{F}(y)-h_{E}(y)$; thus $h_{F\setminus E}=h_{F}+(-1)h_{E}$ is measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. ($\lambda$3) Let $(E^{m})_{m\in\mathbb{N}}$ be a sequence in $\mathcal{D}$ with $E^{m}\subseteq E^{m+1}$ for every $m$, and let $E=\bigcup_{m}E^{m}$, which lies in $\mathcal{Y}\otimes\mathcal{Z}$. For each $y$ the sections $(E^{m})_{y}$ are nondecreasing in $m$ with union $E_{y}$, and their measures $h_{E^{m}}(y)$ are real and bounded above by $1$; by Basic Properties of a Measure §continuity-below the sequence $(h_{E^{m}}(y))_{m\in\mathbb{N}}$ converges to $h_{E}(y)$. Hence $h_{E}$ is measurable by claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $E\in\mathcal{D}$.

By Dynkin's Pi-Lambda Theorem, $\sigma(\mathcal{P})\subseteq\mathcal{D}$, and $\sigma(\mathcal{P})=\mathcal{Y}\otimes\mathcal{Z}$ by Product Sigma-Algebra; so $h_{E}$ is measurable for every $E\in\mathcal{Y}\otimes\mathcal{Z}$. Since $|h_{E}|\le1$ and $\mu(Y)=1$, (P1) shows that $h_{E}$ is integrable with respect to $\mu$. This proves clause 1.

Step 2: clause 2 (Composite measure). For $E\in\mathcal{Y}\otimes\mathcal{Z}$ we have $(\mu\otimes\kappa)(E)=\int_{Y}h_{E}\,d\mu$, which by (P2) is also the integral of $h_{E}$ as a nonnegative measurable function; since $0\le h_{E}\le\mathbf{1}_{Y}$, Linearity and Monotonicity of the Lebesgue Integral §nonnegative and The Integral of an Indicator Function is the Measure of the Set give $0\le(\mu\otimes\kappa)(E)\le\mu(Y)=1$, so $\mu\otimes\kappa$ takes values in $[0,\infty]$. Since $\kappa(y,\varnothing)=0$ for every $y$, $h_{\varnothing}$ is the zero function $0\cdot\mathbf{1}_{Y}$, whose integral is $0\cdot\mu(Y)=0$ by Linearity and Monotonicity of the Lebesgue Integral §nonnegative, The Integral of an Indicator Function is the Measure of the Set and (P2); so $(\mu\otimes\kappa)(\varnothing)=0$.

Countable additivity. Let $(E^{m})_{m\in\mathbb{N}}$ be pairwise disjoint members of $\mathcal{Y}\otimes\mathcal{Z}$ with union $E$. For each $y\in Y$ the sections $(E^{m})_{y}$ are pairwise disjoint members of $\mathcal{Z}$ with union $E_{y}$, so by the countable additivity of the measure $\kappa_{y}$ (Measure, Measure Space, and Probability Measure), $h_{E}(y)=\sum_{m\in\mathbb{N}}h_{E^{m}}(y)$; as this sum is the real number $h_{E}(y)$, by the definition of the sum in Measure, Measure Space, and Probability Measure it is the least upper bound of the partial sums $H_{n}(y)=\sum_{m=1}^{n}h_{E^{m}}(y)$. The functions $H_{n}$ are nonnegative and measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, and $H_{n}\le H_{n+1}$. By Monotone Convergence Theorem on $(Y,\mathcal{Y},\mu)$,

$$\int_{Y}h_{E}\,d\mu=\sup_{n}\int_{Y}H_{n}\,d\mu ,$$

and by the additivity in Linearity and Monotonicity of the Lebesgue Integral §nonnegative, applied $n-1$ times, $\int_{Y}H_{n}\,d\mu=\sum_{m=1}^{n}\int_{Y}h_{E^{m}}\,d\mu=\sum_{m=1}^{n}(\mu\otimes\kappa)(E^{m})$ (all integrals here being nonnegative integrals, equal to the Lebesgue integrals by (P2) and Step 1). Hence the partial sums of $((\mu\otimes\kappa)(E^{m}))_{m\in\mathbb{N}}$ are real, bounded above by the real number $(\mu\otimes\kappa)(E)$, and have least upper bound $(\mu\otimes\kappa)(E)$; by Measure, Measure Space, and Probability Measure this says $(\mu\otimes\kappa)(E)=\sum_{m\in\mathbb{N}}(\mu\otimes\kappa)(E^{m})$. Thus $\mu\otimes\kappa$ is a measure on $(Y\times Z,\mathcal{Y}\otimes\mathcal{Z})$.

Since $h_{Y\times Z}=\mathbf{1}_{Y}$, The Integral of an Indicator Function is the Measure of the Set and (P2) give $(\mu\otimes\kappa)(Y\times Z)=\mu(Y)=1$, so $\mu\otimes\kappa$ is a probability measure. The rectangle formula of clause 2 is (1) integrated against $\mu$.

Step 3: clause 3 (Rectangles determine it). Let $\pi$ be as in clause 3. By Step 2, $\pi$ and $\mu\otimes\kappa$ agree on every member of $\mathcal{P}$. Taking $A=Y$ and $B=Z$ gives $\pi(Y\times Z)=\int_{Y}\mathbf{1}_{Y}(y)\kappa(y,Z)\,\mu(dy)=1=(\mu\otimes\kappa)(Y\times Z)<\infty$. By Step 1, $\mathcal{P}$ is a $\pi$-system, and its generated $\sigma$-algebra in the sense of Generated Sigma-Algebra is $\mathcal{Y}\otimes\mathcal{Z}$ by Product Sigma-Algebra. Claim 1 of Uniqueness of Finite Measures on a Generating Pi-System and the Density of the Exponential Law, applied to the measurable space $(Y\times Z,\mathcal{Y}\otimes\mathcal{Z})$, the $\pi$-system $\mathcal{P}$ and the measures $\pi$ and $\mu\otimes\kappa$, gives $\pi=\mu\otimes\kappa$.

Step 4: clause 4 (Bounded functions). Let $f:Y\times Z\to\mathbb{R}$ be measurable with respect to $\mathcal{Y}\otimes\mathcal{Z}$, and let $M\ge0$ be real with $|f|\le M$. For each $y\in Y$, $f(y,\cdot)$ is measurable with respect to $\mathcal{Z}$ by claim 3 of Sections of Product-Measurable Sets and Maps Are Measurable, and Insertion Maps into Products Are Measurable and is bounded by $M$, so by (P1) on $(Z,\mathcal{Z},\kappa_{y})$ it is integrable and $g_{f}(y)=\int_{Z}f(y,z)\,\kappa_{y}(dz)$ satisfies $|g_{f}(y)|\le M$. By (P1) on the probability space $(Y\times Z,\mathcal{Y}\otimes\mathcal{Z},\mu\otimes\kappa)$ of Step 2, $f$ is integrable with respect to $\mu\otimes\kappa$; and once $g_{f}$ is known to be measurable with respect to $\mathcal{Y}$, (P1) shows that it is integrable with respect to $\mu$. So it remains to prove, for every bounded measurable $f$, the property

(R) $g_{f}$ is measurable with respect to $\mathcal{Y}$ and $\int_{Y\times Z}f\,d(\mu\otimes\kappa)=\int_{Y}g_{f}\,d\mu$.

Let $\mathcal{H}$ be the set of bounded measurable $f:Y\times Z\to\mathbb{R}$ satisfying (R).

(4a) Indicators. Let $E\in\mathcal{Y}\otimes\mathcal{Z}$. Then $\mathbf{1}_{E}(y,\cdot)=\mathbf{1}_{E_{y}}$, so by The Integral of an Indicator Function is the Measure of the Set and (P2), $g_{\mathbf{1}_{E}}(y)=\kappa(y,E_{y})=h_{E}(y)$, measurable by Step 1; and $\int\mathbf{1}_{E}\,d(\mu\otimes\kappa)=(\mu\otimes\kappa)(E)=\int_{Y}h_{E}\,d\mu$ by the same two results and the definition of $\mu\otimes\kappa$. So $\mathbf{1}_{E}\in\mathcal{H}$.

(4b) Linear combinations. Let $f_{1},f_{2}\in\mathcal{H}$ and $a,b\in\mathbb{R}$. Then $af_{1}+bf_{2}$ is measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and bounded. By Linearity and Monotonicity of the Lebesgue Integral §integrable on each $(Z,\mathcal{Z},\kappa_{y})$, $g_{af_{1}+bf_{2}}=ag_{f_{1}}+bg_{f_{2}}$, which is measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; and the same linearity on $(Y\times Z,\mu\otimes\kappa)$ and on $(Y,\mu)$, where all functions involved are integrable by (P1), gives $\int(af_{1}+bf_{2})\,d(\mu\otimes\kappa)=a\int_{Y}g_{f_{1}}\,d\mu+b\int_{Y}g_{f_{2}}\,d\mu=\int_{Y}g_{af_{1}+bf_{2}}\,d\mu$. So $af_{1}+bf_{2}\in\mathcal{H}$. By (4a), (4b) and induction on the number of terms of the standard representation of Simple Function and Its Integral, every simple function on $(Y\times Z,\mathcal{Y}\otimes\mathcal{Z})$ belongs to $\mathcal{H}$.

(4c) Uniform limits. Let $f$ be bounded and measurable, and suppose there are $f_{m}\in\mathcal{H}$ and reals $\delta_{m}\ge0$, for $m\in\mathbb{N}$, such that $(\delta_{m})$ converges to $0$ and $|f-f_{m}|\le\delta_{m}$ everywhere. For $y\in Y$, linearity (Linearity and Monotonicity of the Lebesgue Integral §integrable) and (P1) with $\kappa_{y}(Z)=1$ give $|g_{f}(y)-g_{f_{m}}(y)|=\bigl|\int_{Z}(f-f_{m})(y,z)\,\kappa_{y}(dz)\bigr|\le\delta_{m}$, so $(g_{f_{m}}(y))_{m}$ converges to $g_{f}(y)$, and $g_{f}$ is measurable by claim 5 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Then $|g_{f}-g_{f_{m}}|\le\delta_{m}$ and $|f-f_{m}|\le\delta_{m}$, and the same two results on the probability spaces $(Y,\mathcal{Y},\mu)$ and $(Y\times Z,\mathcal{Y}\otimes\mathcal{Z},\mu\otimes\kappa)$ give

$$\Bigl|\int f\,d(\mu\otimes\kappa)-\int_{Y}g_{f}\,d\mu\Bigr|\le\Bigl|\int(f-f_{m})\,d(\mu\otimes\kappa)\Bigr|+\Bigl|\int_{Y}(g_{f_{m}}-g_{f})\,d\mu\Bigr|\le2\delta_{m},$$

using $\int f_{m}\,d(\mu\otimes\kappa)=\int_{Y}g_{f_{m}}\,d\mu$. Given a real $\varepsilon>0$, choose $m$ with $2\delta_{m}<\varepsilon$; so the left side is less than every positive real, hence $0$, and $f\in\mathcal{H}$.

(4d) Conclusion. Let $f$ be bounded and measurable with $|f|\le M$. The functions $f^{+}=\max(f,0)$ and $f^{-}=\max(-f,0)$ are measurable by claims 1, 2 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, satisfy $0\le f^{\pm}\le M$, and $f=f^{+}-f^{-}$. By claim 2 of Approximation of Measurable Functions by Simple Functions on the measure space $(Y\times Z,\mathcal{Y}\otimes\mathcal{Z},\mu\otimes\kappa)$ with $K=M$, there are nonnegative simple functions $s_{m}$ with $0\le f^{+}-s_{m}\le(1/2)^{m}$ for every $m$, and $((1/2)^{m})_{m}$ converges to $0$ by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §geometric. By (4b) each $s_{m}\in\mathcal{H}$, so $f^{+}\in\mathcal{H}$ by (4c); likewise $f^{-}\in\mathcal{H}$, and $f\in\mathcal{H}$ by (4b). With the remarks at the start of this step, this proves clause 4, the function $g$ of clause 4 being $g_{f}$.

Step 5: clause 5 (Nonnegative functions). Let $f\ge0$ be measurable. For $n\in\mathbb{N}$ let $f_{n}=\min(f,n)$, measurable by claims 1 and 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions, with $0\le f_{n}\le f_{n+1}\le f$ and $f_{n}\le n$; for each point $p$, $f_{n}(p)=f(p)$ as soon as $n\ge f(p)$, so $f(p)$ is the least upper bound of $\{f_{n}(p):n\in\mathbb{N}\}$. Let $g_{n}=g_{f_{n}}$ be the function of clause 4 for $f_{n}$; by Step 4 it is measurable with respect to $\mathcal{Y}$, it is nonnegative because, by (P2), each $g_{n}(y)$ equals the integral of the nonnegative function $f_{n}(y,\cdot)$ as a nonnegative measurable function; and $g_{n}\le g_{n+1}$ by the monotonicity in Linearity and Monotonicity of the Lebesgue Integral §integrable, and $\int f_{n}\,d(\mu\otimes\kappa)=\int_{Y}g_{n}\,d\mu$. Let $G:Y\to[0,\infty]$ be the pointwise least upper bound in $[0,\infty]$ of $(g_{n})$.

For fixed $y$, the functions $f_{n}(y,\cdot)$ are nonnegative, measurable with respect to $\mathcal{Z}$, nondecreasing in $n$, with least upper bound $f(y,\cdot)$; by Monotone Convergence Theorem on $(Z,\mathcal{Z},\kappa_{y})$ and (P2), the integral of $f(y,\cdot)$ as a nonnegative function is $\sup_{n}g_{n}(y)=G(y)$. By the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §integral, and since $|f(y,\cdot)|=f(y,\cdot)$, the function $f(y,\cdot)$ is integrable with respect to $\kappa_{y}$ exactly when $G(y)<\infty$, that is, when $\{g_{n}(y):n\in\mathbb{N}\}$ is bounded above, that is, when there is $K\in\mathbb{N}$ with $g_{n}(y)\le K$ for all $n$. Hence

$$N=\{y\in Y:G(y)=\infty\}=Y\setminus\bigcup_{K\in\mathbb{N}}\bigcap_{n\in\mathbb{N}}\bigl(Y\setminus\{y:g_{n}(y)>K\}\bigr),$$

and each $\{y:g_{n}(y)>K\}$ belongs to $\mathcal{Y}$ by claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line; so $N\in\mathcal{Y}$ by Sigma-Algebra and Measurable Space. For $y\notin N$, (P2) gives $g(y)=G(y)$, the least upper bound of the $g_{n}(y)$; for $y\in N$, $g(y)=0$. Hence for every real $c$,

$$\{y:g(y)>c\}=\Bigl((Y\setminus N)\cap\bigcup_{n\in\mathbb{N}}\{y:g_{n}(y)>c\}\Bigr)\cup N_{c},$$

where $N_{c}=N$ if $c<0$ and $N_{c}=\varnothing$ if $c\ge0$. This set lies in $\mathcal{Y}$, so $g$ is measurable by claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line. Moreover $g\le G$ everywhere and $g=G$ on $Y\setminus N$.

By Monotone Convergence Theorem on $(Y\times Z,\mathcal{Y}\otimes\mathcal{Z},\mu\otimes\kappa)$ applied to $(f_{n})$, and on $(Y,\mathcal{Y},\mu)$ applied to $(g_{n})$ (which also shows that $G$ is measurable), together with (P2),

$$\int f\,d(\mu\otimes\kappa)=\sup_{n}\int f_{n}\,d(\mu\otimes\kappa)=\sup_{n}\int_{Y}g_{n}\,d\mu=\int_{Y}G\,d\mu ,\tag{2}$$

all three as integrals of nonnegative functions in $[0,\infty]$. By the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §integral, $f$ is integrable with respect to $\mu\otimes\kappa$ exactly when the quantity (2) is finite.

Suppose $f$ is integrable. Then $\int_{Y}G\,d\mu<\infty$, so $\mu(N)=0$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §finite; and $\int_{Y}g\,d\mu\le\int_{Y}G\,d\mu<\infty$ by Linearity and Monotonicity of the Lebesgue Integral §nonnegative, so $g$ is integrable. Conversely, suppose $\mu(N)=0$ and $g$ is integrable. In either case $\mu(N)=0$, so $N$ is null by Null Set of a Measure and $g=G$ almost everywhere by A Property Holding Almost Everywhere; hence The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison gives $\int_{Y}G\,d\mu=\int_{Y}g\,d\mu$, which in the converse case is finite because $g=|g|$ is integrable, so $f$ is integrable by (2). In both cases (2) and (P2) give $\int f\,d(\mu\otimes\kappa)=\int_{Y}g\,d\mu$ as Lebesgue integrals. This proves clause 5.

Step 6: clause 6 (Constant kernels). Suppose $\kappa_{y}=\lambda$ for every $y$. The measures $\mu$ and $\lambda$ are $\sigma$-finite in the sense of Measure, Measure Space, and Probability Measure, as the constant sequences $Y,Y,\dots$ and $Z,Z,\dots$ show, since $\mu(Y)=\lambda(Z)=1$. By Existence and Uniqueness of the Product Measure there is exactly one measure $\mu\otimes\lambda$ on $\mathcal{Y}\otimes\mathcal{Z}$ with $(\mu\otimes\lambda)(A\times B)=\mu(A)\lambda(B)$ for all $A\in\mathcal{Y}$ and $B\in\mathcal{Z}$. By Step 2, $\mu\otimes\kappa$ is a measure on $\mathcal{Y}\otimes\mathcal{Z}$, and for $A\in\mathcal{Y}$, $B\in\mathcal{Z}$, using (P2), the scalar rule of Linearity and Monotonicity of the Lebesgue Integral §nonnegative with the real constant $\lambda(B)\ge0$, and The Integral of an Indicator Function is the Measure of the Set,

$$(\mu\otimes\kappa)(A\times B)=\int_{Y}\lambda(B)\mathbf{1}_{A}\,d\mu=\lambda(B)\,\mu(A)=\mu(A)\lambda(B).$$

By the uniqueness in Existence and Uniqueness of the Product Measure, $\mu\otimes\kappa=\mu\otimes\lambda$.