TITLE: Proving the time-integral of Brownian motion is not Markov
QUESTION [6 upvotes]: Let us assume $(W_t)_{t\geq 0}$ is a Brownian motion with continuous paths. Let us define 
$$Y_t:=\int^t_0 W_s\,ds$$
One knows that $(W_t,Y_t)$ is a Gaussian process with strictly positive density on $\mathbb R^2$ for $t>0$. 
Now we want to show that $Y$ is not a Markov process with respect to $\mathcal F^W_t$. So assume that $W_t$ is a Markov process. Then by the Markov property one has for $0<s\leq t$
$$\mathbb E[Y_t\mid \mathcal F_s^W]=\mathbb E[Y_t\mid Y_s] = Y_s+(t-s)W_s$$
Hence therefore 
$$(t-s)W_s=\mathbb E[Y_t\mid Y_s]-Y_s$$
which makes $W_s$ a $\sigma(Y_s)$-measurable random variable. 
Now I had two things in mind about how the proof should be proceeded and the first one is my favourite
Approach 1.
So this all means that there exists some Borel set $U$ such that
$$\{\omega\ : \ W_s(\omega)\in [0,1]\}=\{\omega \ : \ Y_s(\omega)\in U\}$$
Note that $U\neq\emptyset$ because the set on the LHS has positive measure.  So there is this $u\in U$ for which one has the following:
$$Y_s(\omega)=u \Longrightarrow W_s(\omega)\in [0,1]$$
But that cannot be true since one can certainly find an $\omega$ for which $Y_s(\omega)=u$ but $W_s(\omega)>2$. Basically what I'm saying is that one can find an integral with the same value for a totally different function. But here problem arises as: the set $\{Y_s=u\}$ has measure zero so how should we make sense of this? And what makes one sure that another path of the BM $W$ exist for which $W_s(\omega)>2$? 
Approach 2.
So apparently one has $W_s=g(Y_s)$ for some Borel measurable function $g$. Let 
$$A=\{(x,g(x))\in\mathbb R^2 \ : \ x\in [0,1]\}$$
Then
$$0<\mathbb P(Y_s\in [0,1])=\mathbb P((Y_s,W_s)\in A)$$
But $A$ is a graph of Borel-measurable function so it has zero Lebesgue measure contradicting that we had a positive Lebesgue density. 

Question. Are my approaches plausible? Otherwise, what are needed to fix them ?

To be honest,  I'm more certain in approach 2, although I think that approach 1 can be rescued. I also know that this question can be done through covariance characterization of Gaussian Markov processes. I know there is also a post about that on this site, but I'm not really into that.

REPLY [1 votes]: I have a simple way to rescue approach 1. 
So there was this $U\in\mathcal B$ with positive lebesgue measure for which
$$\{W_s\in [0,1]\}=\{Y_s\in U\}$$ 
Note that $\{W_s\in [-2,-1]\}\cap \{Y_s\in U\}$ has positive probability since $(W_s,Y_s)$ has a positive joint denisty. Therefore
$$0<\mathbb P(W_s\in [-2,-1], Y_s\in U)=\mathbb P(W_s\in [-2,-1]\cap [0,1])=\mathbb P(\emptyset)=0$$
Very simple indeed.
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