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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. – George Lowther Oct 11 '11 at 1:47
@GeorgeLowther: Thank you very much. Would you put this comment as an answer if you have time? I will accept it. – Ilya Oct 11 '11 at 7:02
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 $P_x$? – André Caldas Oct 11 '11 at 14:55
@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 $$P_tf(x) = \mathsf E_x[f(X_t)] :=\int\limits_{\mathcal X}f(y)\mathsf P_x\{X_t\in dy\}$$ – Ilya Oct 11 '11 at 15:01
Is this the same as saying that $$P_t(A) = \int_{\mathcal{X}} P_x(X_t \in A) dx?$$ – André Caldas Oct 11 '11 at 16:44
up vote 1 down vote accepted

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.

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