Probability that Brownian Motion hits a point I have some question regarding the notation from the book on Brownian Motion by Peres and Mörters found online here. They define $\mathbb{P}_x$ to mean a probability measure which makes $(B(t))$ a Brownian motion started at $x$ (p. 36).
In Corollary 2.26 on p. 47 in the book on Brownian Motion they want to show that $\mathbb{P}_x\{y\in B(0,1]\} = 0$ for any $x,y\in \mathbb{R}^2$. If $\Omega$ is the underlying probability space on which the Brownian motion is defined this means that for fixed $x,y\in \mathbb{R}^2$
$$\mathbb{P}\{\omega\in \Omega:y\in B((0,1],\omega)\} = 0$$
where $B$ is a Brownian motion starting at $x$.
They do this by first showing, using their notation, that $\mathbb{P}_x\{y\in B[0,1]\} = 0$ for Lebesgue- almost every $x$ where $y$ is arbitrary. They then say that $\mathbb{P}_{B(\varepsilon)}\{y\in B[0,1]\} = 0$ almost surely where $\varepsilon>0$ is arbitrary.
Now I have some trouble understanding what $\mathbb{P}_{B(\varepsilon)}\{y\in B[0,1]\}$ means.
I read this notation as follows: the event consists of all $\omega$ such that $y\in B([0,1],\omega)$ where $B$ is a Brownian motion starting at $B(\varepsilon,\omega)$. However I don't see how to make sense of this based on previous definitions. Previously Brownian motions start at fixed non-random points in $\mathbb{R}^n$. Furthermore saying that a Brownian motion should start from the point where the Brownian motion hits at $\varepsilon$ is also strange, since in defining the Brownian motion we need to simultaneously know where it hits at $\varepsilon$.
Do they mean that if $\tilde{B}$ is another Brownian motion, then almost surely for $\hat{\omega}$ it holds that
$$\mathbb{P}_{\tilde{B}(\varepsilon,\hat{\omega})}\{\omega: y\in B([0,1],\omega)\} = 0$$
for this I see follows from the previous result. Could someone help me with this?
 A: 
Now I have some trouble understanding what $\mathbb{P}_{B(\varepsilon)}\{y\in B[0,1]\}$ means.

It is nothing but $\mathbb{P}_{x}\{y\in B[0,1]\}|_{x=B(\varepsilon)}$.

I read this notation as follows: the event consists of all $\omega$ such that $y\in B([0,1],\omega)$ where $B$ is a Brownian motion starting at $B(\varepsilon,\omega)$.

No.  The event precisely consists of all $\omega$ such that $y \in B([0,1],\omega)$. For instance, if $B$ is defined canonically (i.e. with $\Omega = C([0,1];\mathbb R^2)$ and $B(\cdot,\omega) =\omega(\cdot)$), then this event is nothing else but the set of functions attaining value $y$ at some point. This has nothing to do with the probability $\mathbb P_{x}$ (actually, probabilities, since there are many of them, and they are mutually singular). Naturally, since the value $\mathbb P_x(A)$ is zero for almost all $x$ and since the law of $B(\varepsilon)$ is absolutely continuous, then $\mathbb P_{B(\varepsilon)}(A) = 0$ almost surely.
Actually, the authors are indeed a bit sloppy here: as I've already mentioned, there are many probability measures, so "almost surely" may a priori sound pretty ambiguous. A diligent way of writing this is "for any $x$, $\mathbb P_{B(\varepsilon)}(A) = 0 \ $ $\ \mathbb P_x$-almost surely". And they are using precisely this when they write further (thanks to the Markov property and homogeneity) that
$$
\mathbb P_x (y \in B ([\varepsilon,1])) = \mathbb E_x \mathbb P_{B(\varepsilon)} (y \in B ([0,1-\varepsilon]))=0.
$$

Some elaboration of the last formula:
$$
\mathbb P_x (y \in B ([\varepsilon,1])) = \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) = y\}) = \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) - B(\varepsilon) + B(\varepsilon) = y\})
$$
Thanks to the ($\mathbb P_x$-)independence of $B(t) - B(\varepsilon)$ and $B(\varepsilon)$, this is equal to
$$
\mathbb E_x \mathbb P_x (\{\exists t\in [\varepsilon,1]: B(t) - B(\varepsilon) + z = y\})\big|_{z = B(\varepsilon)}.
$$
Now using the fact that (under $\mathbb P_x$) $B(t) - B(\varepsilon)$ is, up to a time shift by $\varepsilon$, a Brownian motion starting from $0$, we can further rewrite this as
$$\mathbb E_x \mathbb P_0 (\{\exists s\in [0,1-\varepsilon]: B(s) + z = y\})\big|_{z = B(\varepsilon)} = \mathbb E_x \mathbb P_z (\{\exists s\in [0,1-\varepsilon]: B(s) = y\})\big|_{z = B(\varepsilon)}.
$$
