# Introducing a product measure from a sequence of transition kernels

Given measurable spaces $$(\Omega_k,\mathcal {F_k })_{k=1 } ^n$$, $$P_1$$ a probability measure on $$\Omega_1$$, $$P_2$$ a transition kernel from $$\mathcal {F_1}$$ to $$\mathcal {F_2 }$$,$$P_3$$ a transition kernel from $$\mathcal {F_1}\otimes \mathcal {F_2}$$ to $$\mathcal {F_3}$$ ...$$P_n$$ a transition kernel from $$\mathcal {F_1}\otimes \ldots\otimes \mathcal {F_{n-1 }}$$ to $$\mathcal {F_n } \$$. Here $$\ P_m$$ is a transition kernel from $$\mathcal {F_1}\otimes \ldots \otimes \mathcal {F_{m-1 }}$$ to $$\mathcal F_m$$ means that for every $$B \in \mathcal {F_m } , \ P_m( \ , B)$$ is a measurable function from $$\Omega_1 \times \ldots \times \Omega_{m-1 }$$ to $$[0,1]$$ and for every $$(\omega_1,\ldots,\omega_{m-1 } ) \in \Omega_1 \times \ldots \times \Omega_{m-1 }$$ a probability measure on $$\mathcal {F_m}$$.

I would like to show that $$P(B)=\int P_1(d\omega_1) \int P_2(\omega_1,d \omega_2) \ldots \int 1_B(\omega_1,\ldots,\omega_n)P_n(\omega_1,\ldots,\omega_{n-1 } , d \omega_n)$$ defines a measure on $$\Omega_1 \times \ldots\times \Omega_n$$

This should be done by an argument using induction. That is one shows that $$P$$ is a countably additive set function on the set of measurable rectangles $$B_0\times \ldots\times B_n$$ (and then extend this set function by the usual arguments). To show countable additivity of $$P$$ on measurable rectangles we derive this for $$n=1$$ and then extended the result to an arbitrary number $$n$$ with an argument using induction. And this inductive step is what I need help with.

For the base case $$n=1$$ one do as follows:

First one shows that $$\omega_1 \mapsto \int 1_B(\omega_1,\omega_2)P_2(\omega_1,d \omega _2)$$ is $$\mathcal F_1$$ measurable and this is done exactly as usually done in a first step in Fubini's theorem. Then noting that $$B \mapsto \int P_1(d \omega_1)\int 1_B(\omega_1,\omega_2)P_2(\omega_1,d \omega _2)$$ is nonnegative and monotone all we need to verify is the $$\sigma$$-additivity. This follows from the monotone convergence theorem:

\begin{align*}\int P_1(d \omega_1)\int 1_{\cup_{k=1}^{\infty }F_k} (\omega_1,\omega_2)P_2(\omega_1,d \omega _2) &=\int P_1(d \omega_1)\sum_{k=1 }^ {\infty } \int 1_{F_k} (\omega_1,\omega_2)P_2(\omega_1,d \omega _2) \\ &=\sum_{k=1 }^ {\infty }\int P_1(d \omega_1) \int 1_{F_k} (\omega_1,\omega_2)P_2(\omega_1,d \omega _2). \end{align*}

And the inductive step should be something like: assume that $$Q(B)=\int P_1(d\omega_1) \int P_2(\omega_1,d \omega_2)\ldots\int 1_B(\omega_1,\ldots,\omega_{n-1 } )P_{n-2 } (\omega_1,\ldots,\omega_{n-2 } , d \omega_{n-1 } )$$ defines a measure on $$\Omega_1 \times \ldots\times \Omega_{n-1 }$$ then

\begin{align*} & \int P_1(d\omega_1) \int P_2(\omega_1,d \omega_2)\ldots\int 1_{B_0 \times \ldots \times B_n}(\omega_1,\ldots,\omega_n)P_n(\omega_1,\ldots,\omega_{n-1 } , d \omega_n) \\ &=\int 1_{B_0 \times \ldots \times B_{n-1}} \left(\int 1_{B_n } P_n(\omega_1,\ldots,\omega_{n-1 } , d \omega_n) \right) Q(d \omega_1,\ldots,d \omega_{n-1 } ) \end{align*} And we would then be in a similar situation as in the base case.

My question is: If this is correct, how is the equality motivated? Or if not, how should the inductive step be?

Let's start with a technical lemma which we will use several times:

Lemma. Let $$(\Omega_i,\mathcal{F}_i)$$, $$i=1,2$$,be measurable spaces, and let $$P$$ a transition kernel from $$\mathcal{F}_1$$ to $$\mathcal{F}_2$$. If $$X:(\Omega_1 \times \Omega_2, \mathcal{F}_1 \otimes \mathcal{F}_2) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is a bounded measurable mapping then $$(\Omega_1,\mathcal{F}_1) \ni \omega_1 \mapsto \int_{\Omega_2} X(\omega_1,\omega_2) \, P(\omega_1,d\omega_2) \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is measurable.

Idea of proof: For $$X=1_A 1_B$$ with $$A \in \mathcal{F}_1$$ and $$B \in \mathcal{F}_2$$ the assertion is immediate and then we can extend it using a classical monotone class argument.

For measurable spaces $$(\Omega_i,\mathcal{F}_i)$$, $$i \in \{0,\ldots,n\}$$, and transition kernels $$P_i$$ from $$\mathcal{F}_0 \otimes \ldots \otimes \mathcal{F}_{i-1}$$ to $$\mathcal{F}_i$$ set

$$Q_n(\omega_0,B) := \int P_1(\omega_0,d\omega_1) \int P_2(\omega_0,\omega_1,d\omega_2) \ldots \int 1_B(\omega_1,\ldots,\omega_n) \, P_n(\omega_0,\ldots,\omega_{n-1},d\omega_n)$$

Claim: For each $$n \in \mathbb{N}$$ it holds for any choice of measurable spaces and transition kernels that $$Q_n$$ is a transition kernel from $$\mathcal{F}_0$$ to $$\mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_n$$.

Base: Since $$Q_1(\omega_0,B) = P_1(\omega_0,B)$$, the assertion is obvious for $$n=1$$.

Inductive step: Assume that the assertion holds for some $$n \in \mathbb{N}$$. Let $$(\Omega_k,\mathcal{F}_k)$$, $$k \leq n+1$$, be measurable spaces, and $$P_i$$ transition kernels from $$\mathcal{F}_0 \otimes \ldots \otimes \mathcal{F}_{i-1}$$ to $$\mathcal{F}_i$$. For fixed $$(\omega_0,\omega_1) \in \Omega_0 \times \Omega_1$$ consider

\begin{align*} &U((\omega_0,\omega_1),A) \\ &:= \int P_2(\omega_0,\omega_1,d\omega_2) \int P_3(\omega_0,\omega_1,\omega_2,d\omega_3) \ldots \int 1_{A}(\omega_2,\ldots,\omega_{n+1}) \, P_{n+1}(\omega_0,\ldots,\omega_n,d\omega_{n+1}) \end{align*}

for $$A \in \mathcal{F}_2 \otimes \ldots \otimes \mathcal{F}_{n+1}$$. Since $$P_i(\omega_0,\omega_1,\cdot)$$, $$i \in \{2,\ldots,n+1\}$$ are transition kernels from $$\mathcal{F}_2 \otimes \ldots \otimes \mathcal{F}_{i-1}$$ to $$\mathcal{F}_i$$, it follows from our induction hypothesis that $$U$$ is a transition kernel from $$\mathcal{F}_0 \otimes \mathcal{F}_1$$ to $$\mathcal{F}_2 \otimes \ldots \times \mathcal{F}_{n+1}$$. By the above lemma, this implies that $$(\Omega_0 \times \Omega_1, \mathcal{F}_0 \otimes \mathcal{F}_1) \ni (\omega_0,\omega_1) \mapsto \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_A(\omega_0,\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) \tag{1}$$ is measurable for any $$A \in \mathcal{F}_0 \otimes \ldots \otimes \mathcal{F}_{n+1}$$. In particular, $$Q_{n+1}(\omega_0,B)$$ is well defined and

\begin{align*} Q_{n+1}(\omega_0,B) &= \int_{\Omega_1} \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_B(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) \, P_1(\omega_0,d\omega_1) \tag{2} \end{align*}

for any $$B \in \mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_{n+1}$$ and $$\omega_0 \in \Omega_0$$.

Now let's show that $$Q_{n+1}$$ has all the desired properties. Let $$(B_j)_{j \in \mathbb{N}} \subseteq \mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_{n+1}$$ be pairwise disjoint sets and set $$B:= \bigcup_{j \geq 1} B_j$$. Since $$1_B = \sum_{j \geq 1} 1_{B_j}$$ it follows from the monotone convergence theorem that

$$\int_{\Omega_2 \times \ldots \times \Omega_n} 1_B(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) = \sum_{j \geq 1} \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_{B_j}(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega})$$

for any $$(\omega_0,\omega_1) \in \Omega_0 \times \Omega_1$$; note that we can apply the monotone convergence theorem because we have already shown that $$U((\omega_0,\omega_1),\cdot)$$ is a measure. Integrating both sides with respect to $$P_1$$ we find from $$(2)$$ that

$$Q_{n+1}(\omega_0,B) = \int_{\Omega_1} \sum_{j \geq 1} \left( \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_{B_j}(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) \right) P(\omega_0,d\omega_1).$$

Applying once more the monotone convergence theorem we get

\begin{align*} Q_{n+1}(\omega_0,B) & = \sum_{j \geq 1} \int_{\Omega_1}\left( \int_{\Omega_2 \times \ldots \times \Omega_{n+1}} 1_{B_j}(\omega_1,\tilde{\omega}) \, U((\omega_0,\omega_1),d\tilde{\omega}) \right) P(\omega_0,d\omega_1) \\ &= \sum_{j \geq 1} Q_{n+1}(\omega_0,B_j). \end{align*}

This proves the $$\sigma$$-addivity of $$Q_{n+1}(\omega_0,\cdot)$$. Finally, we note that it follows from $$(1)$$, $$(2)$$ and the above lemma that $$(\Omega_0,\mathcal{F}_0) \ni \omega \mapsto Q_{n+1}(\omega_0,B)$$ is measurable for any $$B \in \mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_{n+1}$$. Consequently, $$Q_{n+1}$$ is a transition kernel from $$\mathcal{F}_0$$ to $$\mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_{n+1}$$.

Choosing $$\Omega_0$$ as a trivial space (e.g. consisting of a single element), it follows immediately that

$$P_n(B) = \int P_1(d\omega_1) \int P_2(\omega_1,d\omega_2) \ldots \int 1_B(\omega_1,\ldots,\omega_n) \, P_n(\omega_1,\ldots,\omega_{n-1},d\omega_n)$$

is a measure on $$(\Omega_1 \times \ldots \times \Omega_n,\mathcal{F}_1 \otimes \ldots \otimes \mathcal{F}_n)$$ for any probability measure $$P_1$$ on $$(\Omega_1,\mathcal{F}_1)$$ and any transition kernels $$P_i$$.

• Thank you! Clear and detailed as ever! I have a small question for the countable additivity, how do you get $Q_{n+1}(\omega_0,\cup _{k=1 } ^{\infty} B^{(k) } ) = \int \sum _{k=1 } ^{\infty } 1_{B^{(k)} _1}(\omega_1) U((\omega_0,\omega_1),B_2 ^{(k) } \times \ldots \times B_{n+1}^{(k)} ) \, P_1(\omega_0,d\omega_1).$? On which we then could apply the monotone convergence theorem. – MrFranzén May 17 '19 at 13:07
• On the other hand when I think about it, could we instead apply the monotone convergence theorem $n+1$ times to "pull" the limit out from the $n+1$ integrals. Here using that for any $k$ and for $(\omega_1,...,\omega_k)$ the map $(\omega_{k+1 } ,...,\omega_{n+1 } )\mapsto \int P_{k+1 } ...\int \sum_{k=1 } ^m 1_{B_1^{(k)} \times ... \times B_{n+1 }^{(k)} } P_{n+1 } (\omega_0,...\omega_n, d \omega_{n+1 } )$ is increasing with $m$. (Or maybe this last claim would need some motivation??) – MrFranzén May 17 '19 at 13:08
• @MrFranzén See my edited answer, I hope it's fine now. Let me know if something seems out of place (... there are still some typos, I guess... sorry already in advance) – saz May 17 '19 at 15:35
• Excellent! There is one minor thing. I believe that we get $(2)$ first for sets of the form $B_1\times C$ for sets $B_1 \in \mathcal {F_1 }$ and $C \in \mathcal {F_2} \times ... \times \mathcal {F_{n+1 } }$, since we have to use $1_{B_1 \times C } = 1_{B_1 } 1 _C$ and then we extend the class of sets for which the equation holds using a standard $\pi - \lambda$ argument. – MrFranzén May 18 '19 at 6:35
• And lots of thanks again! – MrFranzén May 18 '19 at 6:37