Lower semicontinuity of the indicator function: stochastic processes Let $X$ be a Markov process given on a metric space $\mathcal X$ by a transition semigroup $P_t$ acting on $\mathbb B(\mathcal X)$ - the set of all bounded and Borel measurable functions. Such a function is said to be $\mathcal C$-lower semicontinuous (l.s.c.) if
$$
\mathsf P_x\{\liminf\limits_{t\downarrow 0}f(X_t)\geq f(x)\}=1
$$
for any $x\in \mathcal X$. I wonder under which conditions on $P_t$ a function $1_A(x)$ is l.s.c. for any open $A$? 
Not to be confused with a usual definition of a l.s.c. function which is not based on the processes.
As I understand it means that starting in an open set, with probability one the process stays there for some positive time. If I am not wrong, that holds for any process with cadlag paths since there exists $\lim\limits_{t\downarrow 0}\,\,X_t = x$ so if $x\in A$ - open, then $\lim\limits_{t\downarrow 0}1_A(X_t) = 1$.
 A: First try... :-)
Notice that
$$
  \{\liminf_{t \downarrow 0} f(X_t) \geq f(x)\}
  =
  \bigcap_{n \in \mathbb{N}} \left\{\liminf_{t \downarrow 0} f(X_t) > f(x) - \frac{1}{n}\right\}.
$$
So, since $\left\{\liminf_{t \downarrow 0} f(X_t) > f(x) - \frac{1}{n}\right\}$ decreases when $n \rightarrow \infty$, what you want is that
$$
  P_x\left\{\liminf_{t \downarrow 0} f(X_t) > f(x) - \frac{1}{n}\right\}
  =
  1
$$
for every $n \in \mathbb{N}$.
That is, for every $\varepsilon > 0$,
$$
  P_x\left\{\liminf_{t \downarrow 0} f(X_t) > f(x) - \varepsilon\right\}
  =
  1
$$
So, the condition is equivalent to
$$
  P_x\left(\bigcap_{t > 0} \bigcap_{0 < s < t} \{f(X_s) > f(x) - \varepsilon\}\right)
  =
  1
$$
for every $\varepsilon > 0$.
Now, assuming $f = 1_A$, we have that if $x \not \in A$,
then the above equation is always true.
For $x \in A$, the condition becomes
$$
  P_x\left(\bigcap_{t > 0} \bigcap_{0 < s < t} \{f(X_s) = 1\}\right)
  =
  1.
$$
But this is the same as
$$
  P_x\left(\bigcap_{t > 0} \bigcap_{0 < s < t} \{X_s \in A\}\right)
  =
  1.
$$
Since the sets $\bigcap_{0 < s < t} \{X_s \in A\}$ increase with $t$,
we can take a sequence $t_j \downarrow 0$, and conclude that the condition
becomes
$$
  P_x\left(\bigcap_{0 < s < t} \{X_s \in A\}\right)
  \uparrow
  1,
$$
when $t \downarrow 0$.
I don't know how to pass from this to $P_t$, since I am not
familiar with Markov processes.
But I guess that you might be able to conclude what @George Lowther
said in his comment: the process is right-continuous.
