There are several versions of the Blumenthal's 0-1 law and proofs of it. Many of them are only done for Brownian Motions. I'm doing a self-study in Stochastic processes and found a version as follows:
if $A\in \mathcal{F}_{0^+}^X$ then $\mathbb{P}_x(A)=0$ or $1$ for all $x\in E$.
To give some background info: $\mathcal{F}_t^X$ is the natural filtration, $x$ is an arbitrary element of $E$, and $\mathcal{F}_{t^+}$ is the germ $\sigma$-algebra namely $\cap_{s>t} \mathcal{F}_s^X$. For more info see http://web.archive.org/web/20170215074818if_/http://www.math.leidenuniv.nl/~spieksma/colleges/sp-master/sp-hvz1.pdf
It should follow from a strong Markov-property version:
Let $\tau$ be an $\mathcal{F}_t^X$ optional time (that is, an $\mathcal{F}_{t^+}$ stopping time), and $Y$ a bounded $\mathcal{F}_{\tau^+}$ measurable rv, $\nu$ initial distribution. Then $$ \mathbb{E}_{\nu} Y f(X_{\tau+t}) = \mathbb{E}_{\nu} (Y \mathbb{E}_{X_{\tau}} f(X_t)). $$
Evaluating $\nu=x$, $Y=1_A$, $\tau=0$, $f=1_B$ with $B\subseteq E$ not yet specified yields $\mathbb{P}(A\cap \lbrace X_t \in B\rbrace) = \mathbb{P}(A)\mathbb{P}(X_t \in B)$, so here we see independence. From this I can't conclude that $A\in \mathcal{F}_t^X$, but if this would be true then $A$ would be independent of itself whence $\mathbb{P}(A) = \mathbb{P}(A)\mathbb{P}(A)$ would follow.
Is there a way to show from (this version of ) the strong Markov property that $A\in\mathcal{F}_{t^+}^X$?
Thanks for your help in advance.
This is not really about the strong Markov property. It's a consequence of the simple Markov property with respect to $(\mathcal F_{t+})$, used only at time $t=0$: $$ \Bbb P_x[A]=\Bbb E_x[1_A\cdot1_A]=\Bbb E_x[1_A\cdot\Bbb P_{X_0}[A]]=\Bbb E_x[1_A\cdot\Bbb P_{x}[A]]=\Bbb E_x[1_A]\cdot\Bbb P_x[A]. $$