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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 $\lim\limits_{t\downarrow 0}1_A(X_t) = 1$.

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    @AndréCaldas that's quite a comeback :) no I don't, but I think you can just google for Markov semigroup – 2016-10-04

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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.