Markov property of Brownian motion

First suppose t>0 and put Y=B_{s+t}-B_s. For any finite set of past times in [0,s], form their increasing union with {0,s,s+t}, delete repetitions, and apply [F1]. The past values differ almost surely from the cumulative sums of the increments up to s only by B_0=0; the final increment Y is independent of all those earlier increments by [F2]. For any Borel past cylinder its indicator agrees almost surely with the corresponding cylinder in those cumulative sums, so its joint probability with {Y in C} factors for every Borel C. This includes cylinders involving time zero and the case s=0, where all past cylinders have probability zero or one. Finite past cylinders generate the entire raw past; the pi-system criterion [F2] therefore makes sigma(Y) independent of that past. Finally Y has law N(0,t), also the law of B_t since B_0=0 almost surely.

Apply [F3] with $X=B_s$, which is ${\mathcal{F}}_s^0$-measurable, with $Y$ as in step 1.1, and with $h(x,y):=f(x+y)$: the function $H(x)={\int }_{\mathbb{R}}f(x+y)\mu (dy)$ with $\mu =N(0,t)$ is Borel, and $E[f(B_{s+t})\mathrm{\mid }{\mathcal{F}}_s^0]=H(B_s)$ almost surely. Since $\mu$ is also the law of $B_t$, step 1.1 and [F4] identify $H(x)=E[f(x+B_t)]=P_{t}f(x)$ for every $x$. This proves the raw-filtration assertion.

Suppose $t>0$ and put $u_n=s+t\mathrm{/}(n+2)$ for integers n>=0, so $u_n\downarrow s$ with $s<u_n<s+t$. Put $h_n=s+t-u_n>0$. Applying step 2.1 at the time pair $(u_n,h_n)$ and then completing the conditioning sigma-algebra as in [F6] gives $E[f(B_{s+t})\mid \bar{\mathcal{F}}{}_{u_n}^0]=P_{h_n}f(B_{u_n})$ almost surely. Indeed, bounded event integrals are unchanged when an event is replaced by a raw event differing by a null set.

On the probability-one continuity event, $B_{u_n}\to B_s$ and $h_n\to t$. The Gaussian densities $v\mapsto p_{h_n}(B_{u_n},v)$ converge pointwise to $p_t(B_s,v)$ and, for all large $n$, are bounded by an integrable envelope: choose a finite M bounding all the centers including B_s. Since $h_n\in [t\mathrm{/}2,t]$, each density is at most $(\pi t{)}^{-1\mathrm{/}2}\exp (-((\mathrm{\mid }v\mathrm{\mid }-M{)}_{+}{)}^2\mathrm{/}(2t))$, which is integrable (bounded on [-M,M], with Gaussian tails). This bound may depend on the fixed path; that is sufficient for this pathwise integral limit. Dominated convergence therefore gives their $L^1$ convergence, and hence $P_{h_n}f(B_{u_n})\to P_{t}f(B_s)$ for bounded $f$. If $A\in {\mathcal{F}}_s={\bigcap }_{u>s}\bar{\mathcal{F}}{}_u^0$, then $A\in \bar{\mathcal{F}}{}_{u_n}^0$ for every $n$, so step 3.1 gives ${\int }_{A}f(B_{s+t})dP={\int }_A{P}_{h_n}f(B_{u_n})dP$. A second bounded dominated-convergence passage yields ${\int }_{A}f(B_{s+t})dP={\int }_A{P}_{t}f(B_s)dP$. Since $P_{t}f(B_s)$ is ${\mathcal{F}}_s$-measurable, [F5] proves the usual-filtration assertion.

For $t=0$ the transition convention gives $P_{0}f=f$ and both identities read $E[f(B_s)\mathrm{\mid }\mathcal{G}]=f(B_s)$, true because $B_s$ is $\mathcal{G}$-measurable for $\mathcal{G}\in \{{\mathcal{F}}_s^0,{\mathcal{F}}_s\}$. The proof also covers $s=0$ when $t>0$, and then $P_{t}f(B_0)=P_{t}f(0)$ almost surely. If $f\equiv 0$ both sides vanish. AC is inherited from the Brownian and normal-law interfaces, the null-envelope witnesses in [F6], and in particular [F7] for conditional expectation; the independence and transition computations make no further choice.