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Let $E$ be a separable Banach space, with Borel $\sigma$-algebra $\mathcal B_E$. Let $X=(X_t)$ be an $E$ valued stochastic process on some filtered probability space $(\Omega,\mathcal F,(\mathcal F_t), \mathbb P)$. Finally assume that $X$ has stationary independent increments, and is right continuous. Now I would like to show that $X$ is necessarily a Markov process with initial distribution $\mu_0(A)=\mathbb P(X_0\in A)$. I know that this is definitely not the first time this has been asked on this site as this, this, and this attest to. However, none of these answer the doubt I have.

I am working with Markov processes defined in terms of transition functions, so in order to show that $X$ is a Markov process I first need to find a suitable transition function. By tansition function I mean a semigroup of transition kernels $(P_t)$ .

My best guess (backed up by Googling) is that for each $t$ we can define $P_t:E\times \mathcal B_E\to [0,1]$ by $P_t(x,A)=\mathbb P(X_t+x\in A)=\mu_t(A-x)$, where $\mu_t$ is the law of $X_t$. Now quite clearly each $P_t(x,\cdot)$ is a probability measure because translations are homeomorphisms, but I am struggling to show that each $P_t(\cdot,A):E\to [0,1]$ is $\mathcal B_E/\mathcal B_{[0,1]}$ measurable, which is needed for each $P_t$ to be a transition kernel. For any $t\in \mathbb R^+$, $A\in \mathcal B_E$ and $(a,b)\subset [0,1]$ we have that $P_t(\cdot,A)^{-1}(a,b)=\{x\in E:\mathbb P(\mathbb X_t+x\in A)\in (a,b)\}$. If we knew that $\mu_t$ was translation invariant (impossible in an infinite dimensional space) then it would be trivial to show that this set is measurable, but alas life is not so easy. I'm guessing that there is some simple continuity kind of argument around translations and distributions that I'm missing, but as I'm missing it its simplicity to me is not helpful at the moment. I've been mulling over this for a few days with no success, so would appreciate any help in proving the measurability of $P_t(\cdot,A)$.

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Lemma Let $\nu$ be a probability measure on $\mathcal{B}(E)$. Then $$x \mapsto \int_E 1_C(x,y) \, \nu(dy) \tag{1}$$ is $\mathcal{B}(E)/\mathcal{B}(\mathbb{R})$-measurable for any $C \in \mathcal{B}(E) \otimes \mathcal{B}(E)$.

Proof: The statement is clearly true for any set $C$ of the form $C=A \times B$ where $A,B \in \mathcal{B}(E)$. If we set

$$\mathcal{D} := \{C \in \mathcal{B}(E) \otimes \mathcal{B}(E); \text{(1) is Borel measurable}\}$$

this means that $\mathcal{B}(E) \times \mathcal{B}(E) \subseteq \mathcal{D}$. If we can show that $\mathcal{D}$ is a Dynkin system, it follows that $\mathcal{D} = \mathcal{B}(E) \otimes \mathcal{B}(E)$.

  • $E \times E \in \mathcal{D}$: That's obvious.
  • stable under complement: Let $C \in \mathcal{D}$. Since $1_{C^c} = 1_{E \times E}-1_C$ we find that $$x \mapsto \int_E 1_{C^c}(x,y) \, \nu(dy) = \int_{E} \nu(dy) - \int 1_C(x,y) \, \nu(dy)$$ is $\mathcal{B}(E)/\mathcal{B}(\mathbb{R})$-measurable, and so $C^c \in \mathcal{D}$.
  • stable under disjoint unions: Let $(C_n)_{n \in \mathbb{N}} \subseteq \mathcal{D}$ be pairwise disjoint. Using the monotone convergence theorem we find for $C:= \bigcup_{n \in \mathbb{N}} C_n$ that $$\int 1_C(x,y) \, \nu(dy) = \sum_{n \geq 1} \int 1_{C_n}(x,y) \, \nu(dy) =\lim_{N \to \infty} \sum_{n =1}^N \int 1_{C_n}(x,y) \, \nu(dy)$$ is Borel measurable (with respect to $x$), and so $C \in \mathcal{D}$.

Applying the above result for the distribution $\nu := \mathbb{P}(X_t \in \cdot)$ we find that

$$x \mapsto \int 1_C(x,y) \, \nu(dy) = \mathbb{E}(1_C(x,X_t))$$

is measurable for any $C \in \mathcal{B}(E) \otimes \mathcal{B}(E)$. Choosing

$$C := \{(x,y) \in E \times E; x+y \in A\}$$

for some given set $A \in \mathcal{B}(E)$, this proves the assertion.

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  • $\begingroup$ This is exactly the missing thing I needed. Thank you $\endgroup$ – K.Power May 9 '18 at 13:02
  • $\begingroup$ @K.Power You are welcome. $\endgroup$ – saz May 9 '18 at 16:59

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