If we consider a Brownian motion starting at large $$ say $x=10^8$ i.e $B_0=x \text{ -}P^{x}$ almost surely then why $E^x[B_{\epsilon}]=0$ for $\epsilon$ very small. Wouldn't the expectation be close to $x$. Is it zero beacuse the measure $P^x$ assigns lower weights to values $B_{\epsilon}(\omega)$ near $x$.
It seems a bit Bm has continuous paths.
I apologize if the question is not well formulated.
If your Brownian motion starts from $x$, then at time $\varepsilon > 0$ we have that $B_\varepsilon \sim N(x, \varepsilon)$. In particular $0 \neq \mathbb{E}^x[B_\varepsilon] = x$.