$ 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.



Continuity in intial state of Brownian Motion
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.


