An example of a process which is markovian but it is not strongly markovian I need to show that the following process is markovian but not strongly markovian.

Consider the following process $(X_{t})_{\mathbb{R}_{+}}$, starting from
  position $x$
  
  
*
  
*if $x = 0$ then $X_t=0$ for all times.
  
*if $x \not = 0$ then $X$ is a standard Brownian motion, say $W_{t}$ starting from $x$.
  

I need to show that this process is markovian but not strongly markovian.
Firstly, I know that the transition function of $(X_{t})$ is given by
$$p_{t}f(x)= \left \lbrace \begin{array}{cc}  b_{t}f(x), & \text{ if } x \neq 0\\
f(0), & \text{ if } x=0 \end{array} \right. $$
where $b_{t}$ is the transition function of the brownian motion $W_{t}$.
And I am stuck in proving the markovianity, i.e. I don't know how to prove that, for any $B$ a Borel set we have that
$$E[\mathbb{1}_{B}(X_{t+s})|\mathcal{F}_{s}]=E[\mathbb{1}_{B}(X_{t+s})|X_{s}]$$ 
If anyone could help me with this I would be very thankful.
 A: Let's consider separately the case $x=0$ and $x \neq 0$.

Case 1: Starting point $x=0$

If $x=0$, then $X_t=0$ for all $t \geq 0$, and so
$$\mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) = \mathbb{E}(1_B(0) \mid \mathcal{F}_s) = 1_B(0)$$ for all $s,t \geq 0$. Exactly the same reasoning gives
$$\mathbb{E}(1_B(X_{t+s}) \mid X_s) = \mathbb{E}(1_B(0) \mid X_s) = 1_B(0).$$ Hence, $$\mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) = \mathbb{E}(1_B(X_{t+s}) \mid X_s).$$

Case 2: Starting point $x \neq 0$

If $x \neq 0$, then $X_t=x+W_t$ for some Brownian motion $(W_t)_{t \geq 0}$ started at $0$. Note that the canonical filtration of $(X_t)_{t \geq 0}$ is the canonical filtration of $(W_t)_{t \geq 0}$, i.e. there is no need to distinguish between the two filtrations. By the Markov property of Brownian motion, we have
\begin{align*} \mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) &= \mathbb{E}(1_B(x+W_{t+s}) \mid \mathcal{F}_s) = g(W_s) \end{align*} for $$g(y) := \mathbb{E}(1_B(x+y+W_t)).$$
Equivalently,
$$\mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) = g(W_s+x-x) = g(X_s-x).\tag{1}$$
Since the right-hand side is $\sigma(X_s)$-measurable, it follows from the tower property of conditional expectation that
\begin{align*}\mathbb{E}(1_B(X_{t+s}) \mid X_s) &= \mathbb{E} \big[ \mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) \mid X_s \big] \\ &\stackrel{(1)}{=} \mathbb{E}(g(X_s-x) \mid X_s) = g(X_s-x). \tag{2} \end{align*}
Hence,
$$\mathbb{E}(1_B(X_{t+s}) \mid \mathcal{F}_s) \stackrel{(1)}{=} g(X_s-x) \stackrel{(2)}{=} \mathbb{E}(1_B(X_{t+s}) \mid X_s). $$
