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3 added 4 characters in body

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 $h$ 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$. One nice treatment of these questions of probabilistic potential theory is found in Kai Lai Chung's "Lectures from Markov Processes to Brownian Motion". The lim inf result above is Theorem 3 (p.164) in section 4.4 of this book. 2 I missed a condition in the problem; added 2 characters in body Edit: I just noticed that the OP asked about a 1-d Brownian motion. The constriction below only works in three or more dimensions. Back to the drawing board.... Your function is not necessarily continuous. Its continuity properties depend not only on the function$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$, the function$h$has limit 1. Thus,$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$. One nice treatment of these questions of probabilistic potential theory is found in Kai Lai Chung's "Lectures from Markov Processes to Brownian Motion". The lim inf result above is Theorem 3 (p.164) in section 4.4 of this book. 1 Your function is not necessarily continuous. Its continuity properties depend not only on the function$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$, the function$h$has limit 1. Thus,$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.