Markov property misunderstanding I am reading 'Green, Brown & Probability' by Kai Lai Chung, and on chapter 5 (Markov Property) page 30, it says that the Markov property DOES NOT says that for $T=2$, $P\big(3.14< \lvert X_3 \rvert <3.15 \big| 3<\lvert X_1 \rvert<4 , 3.1 <\lvert X_2 \rvert< 3.2 \big)=P\big(3.14< \lvert X_3 \rvert <3.15 \big| 3.1 <\lvert X_2 \rvert< 3.2\big)$ 'namely that the past $3 < \lvert X_1 \rvert < 4$ may well have an after- effect on the future when the present \X2\ is given as shown'. But I don't see it, is this related to independence of increments?

This has nothing to do with independent increments. The Markov property works when we have an equality assumption for the "present", e.g. $$P(X_3\in A\mid X_2 = x, X_1\in B) = P(X_3\in A\mid X_2 = x). $$ However, when there is a range of values, as in your question, it is possible that the condition on the "past" value $X_1$ makes some particular values of the "present" $X_2$ more probable.

For example, let $X$ be a Poisson process. Consider e.g. the probability $$ P(X_3 \in\\{100,50\\}\mid X_2\in\\{100,1\\}, X_1>50). $$ Clearly, it is equal to $$ P(X_3 \in\\{100,50\\}\mid X_2=100, X_1>50) = P(X_3 =100\mid X_2=100) = e^{-1}. $$ On the other hand, the conditional probability $$ P(X_3 \in\\{100,50\\}\mid X_2\in\\{100,1\\}) $$ is close to $P(X_3 \in\\{100,50\\}\mid X_2=1)$ (since $X_2=100$ is highly unlikely) and is extremely small. Consequently, $$ P(X_3 \in\\{100,50\\}\mid X_2\in\\{100,1\\}) \
eq P(X_3 \in\\{100,50\\}\mid X_2\in\\{100,1\\}, X_1>50). $$