Consider the stochastic process $(X_t)_{t \in \mathbb{R}}$ and show the equivalence of the following two Markov properties:
(a) $P(X_t \in A \mid X_u, u \leq s) = P(X_t \in A\mid X_s) \qquad \forall A \in \mathcal{S}, \forall t>s \in \mathbb{R}$;
(b) $P(A\cap B\mid X_t) = P(A\mid X_t)P(B\mid X_t) \text{ almost surely} \qquad \forall t \in \mathbb{R}, \forall A \in \mathcal{F}_t, \forall B \in \mathcal{T}_t$.
"The future and the past are independent given the present."
As suggested I try to show the discrete case first, where the random variables take discrete values and time is discrete. So far with the given comments and answers:
(b) => (a)
Set $A := \cap_k [X_{k < n} = i_k], B := [X_n+1 = i], C := [X_n = i_n]$, then (b) implies
$P(A\cap B|C) = P(A|C)P(B|C) \iff \frac{P(A\cap B\cap C)}{P(C)} = \frac{P(A\cap C) P(B\cap C)}{P(C)^2} $ $\iff P(B|C) = P(B|A\cap C)$
For these special $A,B,C$ the latter is equivalent to (a) (as we have shown for the discrete case in a previous evercise).
(a) => (b)
Analog/reverse to above reasoning, (a) implies (b) for events $A := \cap_k [X_{k < n} = i_k], B := [X_n+1 = i], C := [X_n = i_n]$ since for such events in above reasoning all steps where biimplications.
It remains to conclude that (a) implies (b) for events of the form $A=\bigcap_i[X_{u_i}=x_i]$, $B=\bigcap_j[X_{s_j}=z_j]$ where $u_i\leqslant n\leqslant s_j$ as well:
Again $P(A\cap B | C) = P(A|C)P(B|C) \iff P(B|C) = P(B|A\cap C)$
so we need to show that the latter holds also for the general events $A$ and $B$ only assuming (a) and the first part we have already shown; and it indeed follows from the first bit: $P(B|A\cap C) = P(\bigcap_j[X_{s_j}=z_j] | A\cap C) = P(\bigcap_j[X_{s_j}=z_j] | C)$ since $s_j > n$.
Is this the discrete case completed or am I missing something? I think somehow I need to handle the special cases where we have one $s_j = n$ or $u_i=n$.
How can I derive the continuous case from the discrete case then? And where does the "almost surely" some in in the continuous case?