Assume $B_t(\omega)=\omega(t),\omega\in (C,\mathcal{C},\mathbb{P}^x)$ is a B.M.(C is the continuous function space ,$\mathcal{C}$ is generated by the coordinate maps)

In Durrett's textbook,he proved the Markov property .

If $s>0$ and $Y:C\to\mathbb R$ is bounded and measuable, then $$\mathbb{E}(Y\circ\theta_s|\mathcal{F_s})=\mathbb{E}_{B_s}Y.\tag{1}$$

here $\mathbb{E}_{B_s}Y$ is the compostion of $\mathbb{E}_{x}Y$ and $B_s$

My question is:

If (1) holds ,then $\mathbb{E}_{B_s}Y$ must be $\mathcal{F_s}$ measurable,$B_s$ is $\mathcal{F_s}$ measurable ,so we must prove that $$x\to \mathbb{E}_x (Y) \text{is measureable}\tag{2}$$

First I prove for bounded and continuous function $Y$,if (2) holds for such $Y$,then (2) holds for $1_F$ where $F$ is a closed set in $C$,then (2) holds for general bounded and measurable $Y$

noticed that $\mathbb E_x(Y(\omega))=\mathbb E_0(Y(\omega +x))$

if $x_n\to x$ in $\mathbb R$ then $\mathbb E_{x_n}(Y(\omega))=\mathbb E_0(Y(\omega+x_n))\to E_0(Y(\omega+x_0))=\mathbb E_{x_0}(Y(\omega))$

"$\to$" is from the continuous of $Y$ and $\omega+x_n\to\omega+x_0$ in the maximum norm.

so $E_x(Y)$ is continuous hence measurable .

but there is a question:$\mathcal C$ is generated by the coordinated maps,does this $\sigma$-algebra equal to the Borel sigma algebta generated by open sets(correspond to the maximum norm)?


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