I have a conjecture that seems rather obvious but the proof seems elusive.

Consider a diffusion given by, $dX_t = \mu(X_t) dt + \sigma(X_t) db_t$

where $b_t$ is a standard Brownian motion.

$\mu,\sigma$ are piecewise Lipschitz, bounded, of locally bounded variation. If there are any other nice properties required, I would assume I have them.

Suppose we are interested in computing the following -

$v(x) = \mathbb{E}^x \int_0^\infty e^{-rt} f(X_t) dt$

where $f$ is the flow payoff that our agent collects. $f$ is also piecewise Lipschitz, bounded, locally bdd variation etc. Let $f$ be between $m$ and $M$.

I want to prove that $v$ is continuous. The reason why I think it sounds obviously true is the following -

Suppose we want to prove continuity at $x$. Fix some point $a$ to the left of $x$ and consider a sequence $\{x_n\} \in [a,x]$ such that $x_n \rightarrow x$.

Let $\tau_n := \inf \{t: X^{x_n}_t \notin (a,x)\}$

Then,

$v(x_n) \le \mathbb{E}^{x_n} \left[ e^{-r\tau_n} [ P(X^{x_n}_{\tau_n} = a) v(a) + P(X^{x_n}_{\tau_n} = x) v(x)] + (1-e^{-r\tau_n})M\right]$

$M$ is because that's the upper bound of $f$.

Now, if, $x_n \rightarrow x$ $\Rightarrow$ $\tau_n \rightarrow 0$, and $P(X^{x_n}_{\tau_n} = x) =1$.

then we can get continuity. But I do not know how to prove these two things which sound so obviously true. The only approach I am aware of is the time change formula but that seems like an overkill for this problem. Any help would be immensely appreciated. Thanks.