Continuity in intial state of Brownian Motion $ B = (B_t, \mathcal{F}_t; t\ge 0 ) $ is a 1-d Brownian family on a
measurable space $(\Omega, \mathcal{F})$ with a family of probability
measures $\{\mathbb{P}^x\}$, i.e. $\mathbb{P}^x(B_0
= x) = 1$, and $B$ is 1-d BM starting from $x$ under $\mathbb{P}^x$.
Let $\tau$ be a given stopping time w.r.t. underlying filtration, $f$ be a given continuous bounded real function. Consider $V(x) = \mathbb{E}^x [f(B_\tau)]$, where
$\mathbb{E}^x$ is the expectation under $\mathbb{P}^x$.
[Question] Is $V(\cdot)$ continuous for any given stopping time
$\tau<\infty$? If not, is there any counter example? Or does continuity
hold with further conditions?
If $\tau$ is deterministic, then $V$ has no doubt to be continuous. I
am not sure, even if the problem is well formulated with the extension
to stopping time $\tau$. Thanks for any of your comments.
 A: Here is a simpler example that I hope convinces you
that $V$ need not be continuous, even in the one dimensional case.
Take one dimensional Brownian motion $(B_t)$
and define the stopping time $\tau(\omega)=1_{(B_0(\omega)<0)}$.
Then, for any bounded measurable $f$, we have
$$V(x)=E_x[f(B_1)]1_{(-\infty,0)}(x)+f(x) 1_{[0,\infty)}(x).$$
The function $V$ can be made discontinuous at zero by choosing $f$ to
have a strict maximum at $x=0$, since then $E_x[f(B_1)] < f(0)$.
Comment:  You really cannot expect the function $V$ to be continuous in general.
The values of a typical stopping time $\tau$ are intimately tied
up with the sample paths of the Brownian motion; in your
words $\tau$ is "strongly correlated'' with $\omega$.
It's in the definition of stopping time.
The only stopping times that are independent of the Brownian motion
are the deterministic ones.
A: Edit: I just noticed that the OP asked about a 1-d Brownian motion.
The constriction below only works in three or more dimensions.
Back to the drawing board....

Your function $V$ is not necessarily continuous. Its continuity
properties depend not only on the function $f$, but also the
nature of the random time $\tau$.
A classic counterexample is found by letting $\tau$ to be 
the hitting time of the complement of a bounded, open region $D$ 
with an irregular point (as defined in Newtonian potential theory). 
For instance, you could choose a region with a "Lebesgue spine".
http://en.wikipedia.org/wiki/Lebesgue_spine
Then $E_x[\tau]<\infty$ for all $x$, and for any continuous $f$
your function $V(x):=E_x[f(B_\tau)]$ is the Perron-Wiener-Brelot
solution to the Dirichlet problem with data $(D,f)$. 
That is, $V$ is harmonic on $D$ and $\lim_{D\ni x\to z}V(x)=f(z)$ at
all regular points $z\in \partial D$. 
However, if the point $z$ is irregular, then choosing $f$ 
with $f(z)=1$ and $f(y)<1$ otherwise, we have $\liminf_{D\ni x\to z}V(x)$<1.
On the other hand, $V(y)=f(y)$ for all $y\not\in \bar D$ so approaching 
the tip of the spine from outside of $\bar D$, the function $V$ has limit 1.
Thus, $V$ fails to be continuous at $z$. Note that $f$ can be as 
smooth as you like.  
Intuitively, the reason why $V$ is discontinuous is that the spine 
is so sharp that Brownian motion fails to see it, even as the starting
point approaches the tip of the spine from within $D$. 
One nice treatment of these questions of probabilistic potential theory
is found in Kai Lai Chung's "Lectures from Markov Processes to Brownian 
Motion". The lim inf result above is Theorem 3 (p.164) in section 4.4 of this book. 
A: Well I am not sure about it, but it could worth a try to start with this to show continuity:
$E^x[f(B_\tau)]=E^0[\int_0^{+\infty}f(x+W_t)dP^{\tau}(t)]=\int_{\mathbb{R}}\int_0^{+\infty}C(t)f(x+y)e^{-\frac{y^2}{2.t}}dP^{\tau}(t)dy$
(where $W_t$ is a BM starting from $0$, $C(t)$ is a normalising constant, and $P^{\tau}(t)$ is the cdf of $\tau$)
Your move now Kenneth 
Regards
