Can we define probability measures $\text P_x$ on $C([0,\infty))$ such that the coordinate process is a Brownian motion started at $x$?

Question

Let $(W_t)_{t\ge0}$ be a continuous Brownian motion on a probability space $(\Omega,\mathcal A,\operatorname P)$ and $$\pi_t:C([0,\infty))\to\mathbb R\;,\;\;\;x\mapsto x(t)$$ for $t\ge0$. Equip $C([0,\infty))$ with the topology of compact convergence. Then $$\mathcal B(C([0,\infty))=\sigma\left(\pi_t,t\ge0\right)\tag1$$ and hence $W:\Omega\to C([0,\infty))$ is $\left(\mathcal A,\mathcal B(\tau)\right)$-measurable.

So, the process $(\pi_t)_{t\ge0}$ is a Brownian motion on $(C[0,\infty),\mathcal B(C[0,\infty)),\operatorname P_0)$, where $\operatorname P_0:=\operatorname P\circ\:W^{-1}$.

Now it is tempting to consider for every $x\in\mathbb R$ a separate continuous Brownian motion started in $x$ and obtain probability measure $\operatorname P_x$ on $(C[0,\infty),\mathcal B(C[0,\infty))$ in this way.

Question: Are we able to do this such that the map $x\mapsto\operatorname P_x(B)$ is Borel measurable for all $B\in\mathcal B(C([0,\infty))$?

Answer 1

If you define $\operatorname P_x$ to be the image of $\operatorname P_0$ under the translation $w\mapsto w+x:=\{w(t)+x, t\ge 0\}$ on $C[0,\infty)$, then $x\mapsto \operatorname P_x$ is Borel measurable, in the sense that $x\mapsto\int F(w)\operatorname P_x(dw)$ is Borel measurable for each bounded $\mathcal B(C[0,\infty))$-measurable function $F$. Indeed, $x\mapsto\int F(w)\operatorname P_x(dw)$ is continuous if $F$ has the form $$ F(w)=f_1(w(t_1))f_2(w(t_2))\dots f_n(w(t_n)), $$ with $n$ a positive integer, $0\le t_1<t_2<\dots <t_n$ and each $f_k$ bounded and continuous. The asserted measurability then follows for general bounded measurable $F$ by a monotone class argument.

Answer 2

Let us step backward for a moment and think about what we really need to obtain the claim:

Step 1

Let $(G,\mathcal G)$ be a measurable group, $$\tau_x:G\to G\;,\;\;\;y\mapsto y+x$$ for $x\in G$, $\mu$ be a probability measure on $(G,\mathcal G)$ and $$\kappa(x,\;\cdot\;):=\tau_x(\mu).$$ By the very definition of a measurable group, $$\theta:G^2\to G\;,\;\;\;(x,y)\mapsto x+y$$ is $(\mathcal G^{\otimes2},\mathcal G)$-measurable. Thus, if $B\in\mathcal G$, then $A:=\theta^{-1}(B)\in\mathcal G^{\otimes2}$ and $$\kappa(\;\cdot\;,B)=\int\mu({\rm d}y)1_A(y,\;\cdot\;)\tag2$$ is $\mathcal G$-measurable by Fubini's theorem. So, $\kappa$ is a Markov kernel on $(G,\mathcal G)$.

Step 2

Now let $(X_t)_{t\ge0}$ be an $(G,\mathcal G)$-valued Lévy process on $(\Omega,\mathcal A,\operatorname P)$ and $$\kappa_t(x,\;\cdot\;):=\mathcal L(X_t+x)\;\;\;\text{for }x\in G\text{ and }t\ge0.$$ If $$\mu_t:=\mathcal L(X_t)\;\;\;\text{for }t\ge0,$$ we see that $$\tau_x(\mu_t)=\mathcal L(X_t+x)\;\;\;\text{for all }t\ge0\tag3$$ and, by step 1, $\kappa_t$ is a Markov kernel on $(G,\mathcal G)$ for all $t\ge0$. (Actually, $(\kappa_t)_{t\ge0}$ is a convolution semigroup on $(G,\mathcal G)$.)

Step 3

Now let $$\rho(x,B):=\operatorname P\left[(X_t)_{t\ge0}+x\in B\right]\;\;\;\text{for }(x,B)\in G\times\mathcal G^{\otimes[0,\:\infty)}.$$

In order to show that $\rho$ is a Markov kernel, let $$\tilde\tau_x:G^{[0,\:\infty)}\to G^{[0,\:\infty)}\;,\;\;\;(y_t)_{t\ge0}\mapsto(y_t+x)_{t\ge0}.$$ As in the question, let $$\pi_t:G^{[0,\:\infty)}\to G\;,\;\;\;y\mapsto y_t$$ for $t\ge0$. Since $\mathcal G^{\otimes[0,\:\infty)}=\sigma(\pi_t,t\ge0)$, for every $x\in G$, the map $\tilde\tau_x$ should be $\mathcal G^{\otimes[0,\:\infty)},\mathcal G^{\otimes[0,\:\infty)})$-measurable if and only if $\tilde\tau_x^{-1}(B)\in\mathcal G^{\otimes[0,\:\infty)}$ for every $B$ of the form $B=(\pi_{t_1},\ldots,\pi_{t_n})^{-1}(A)$, where $A\in\mathcal G^{\otimes[0,\:\infty)}$, $n\in\mathbb N$ and $0\le t_1<\cdots<t_n$. But that's clearly the case, since $\tau_x$ is $(\mathcal G,\mathcal G)$-measurable.

Step 4

In the context of the question, the topology of compact convergence on $\mathbb R^{[0,\:\infty)}$ is metrizable by a metric $d$ and hence $(\mathbb R^{[0,\:\infty)},d)$ is a measurable group. So, step 3 immediately yields the desired result, but with $C([0,\infty))$ replaced by $\mathbb R^{[0,\:\infty)}$.

However, since $\mathcal B(C([0,\infty)),d)=\sigma(\left.\pi_t\right|_{C([0,\:\infty))})$, everything I wrote in step 3 should still hold when $\mathbb R^{[0,\:\infty)}$ is replaced by $C([0,\infty)$.

I would highly appreciate if someone could verify if I made any mistake in my reasoning above!