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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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    Yes cadlag (or, more generally right-continuous) paths gives a positive answer. You do need to choose a good version of the process otherwise $\liminf_{t\downarrow 0}f(X_s)$ will not be measurable. Assuming joint measurability, separability of the metric space and the strong Markov property, then your condition will be true *if and only if* the process is right-continuous.2011-10-11
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    @GeorgeLowther: Thank you very much. Would you put this comment as an answer if you have time? I will accept it.2011-10-11
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    Sorry to make such a basic question, but how does $P_x$ relates to $P_t$. For me, $P_x$ is a probability on $\mathcal{X}^{\mathbb{R}}$... in fact, there is a probability $P$ on $\mathcal{X}^{\mathbb{R}}$, and $P_x = P(\cdot | X_0 = x)$. How does $P_t$ determine $P_x$?2011-10-11
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    @AndréCaldas: sorry for confuse you with symbols: in fact I used different fonts for the probability measure and the semigroup. $$\mathsf P_x = \mathsf P\{\cdot|X_0 = x\}$$ and $$P_tf(x) = \mathsf E_x[f(X_t)] :=\int\limits_{\mathcal X}f(y)\mathsf P_x\{X_t\in dy\}$$2011-10-11
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    Is this the same as saying that $$P_t(A) = \int_{\mathcal{X}} P_x(X_t \in A) dx?$$2011-10-11
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    @AndréCaldas: thank you very much, I go through your answer soon. With regards to $P_t$ you can read about Markov semigroup here: http://www.math.brown.edu/~schhita/Semigroups.pdf where it is denoted in the terms of $T_t$2011-10-11
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    @Ilya: Do you have another link for this semigroup thing?2016-10-04
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    @AndréCaldas that's quite a comeback :) no I don't, but I think you can just google for Markov semigroup2016-10-04

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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.