# If $X$ is a Markov process, can we find a mild assumption ensuring that $\frac1t\operatorname E_x\left[\int_0^tc(X_s)\:{\rm d}s\right]\to c(x)$?

Let

• $$(E,\mathcal E)$$ be a measurable space with $$\{x\}\in\mathcal E$$ for all $$x\in E$$
• $$\mathcal E_b:=\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\}$$
• $$(\kappa_t)_{t\ge0}$$ be a Markov semigroup on $$(E,\mathcal E)$$
• $$(\Omega,\mathcal A)$$ be a measurable space
• $$(X_t)_{t\ge0}$$ be an $$(E,\mathcal E)$$-valued process on $$(\Omega,\mathcal A)$$
• $$\left(\mathcal F^X_t\right)_{t\ge0}$$ denote the filtration generated by $$(X_t)_{t\ge0}$$
• $$\operatorname P_x$$ be a probability measure on $$(\Omega,\mathcal A)$$ with $$\operatorname P_x[X_0=x]=1\tag1$$ and $$\operatorname E\left[f(X_{s+t})\mid\mathcal F^X_s\right]=(\kappa_tf)(X_s)\tag2\;\;\;\text{for all }f\in\mathcal E_b\text{ and }s,t\ge0$$
• $$c:E\to[0,\infty)$$ be $$\mathcal E$$-measurable

Assume that $$X:\Omega\times[0,\infty)\to E$$ is $$(\mathcal A\otimes\mathcal B([0,\infty)),\mathcal E)$$-measurable and hence $$Y_t:=\int_0^tc(X_s)\:{\rm d}s$$ is a well-defined $$[0,\infty]$$-valued random variable on $$(\Omega,\mathcal A)$$ for all $$t\ge0$$.

I'm searching for a mild additional assumption, ensuring that $$\frac{\operatorname E_x[Y_t]}t\xrightarrow{t\to0+}c(x)\tag3.$$ For example, I could imagine that we need to assume that

1. $$E$$ is a topological space and $$\mathcal E=\mathcal B(E)$$
2. $$c$$ is locally bounded
3. $$(X_t)_{t\ge0}$$ is càdlàg

Would this be enough to conclude?

We may note that $$(\kappa_t)_{t\ge0}$$ is a contraction semigroup on $$\mathcal E_b$$. If $$(\kappa_t)_{t\ge0}$$ would be strongly continuous, we could conclude that $$\frac1t\int_0^t\kappa_sf\:{\rm d}s\xrightarrow{t\to0+}f\tag4$$ for all $$f\in\mathcal E_b$$. However, $$(\kappa_t)_{t\ge0}$$ doesn't need to be strongly continuous and hence we cannot apply this. On the other hand, $$(4)$$ is clearly stronger than what we need for $$(3)$$ to hold.

To begin with, we might want to note that $$c_n:=\min(c,n)\in\mathcal E_b\;\;\;\text{for all }n\in\mathbb N.$$

• The simple way is to assume that $X$ is right-continuous, that $c$ is continuous at $x$ and bounded. Then Lebesgue dominated convergence theorem applies after a change of variable $s=tu$. Jul 5 at 19:53
• @ChristopheLeuridan Thank you for your comment. Yes, I know, but unfortunately boundedness of $c$ is a too strong assumption. The only thing I'm willing to admit in that direction is local boundedness of $c$. Jul 6 at 5:07
• @ChristopheLeuridan Let $Z:=c\circ X$. After we have shown that, as long as $c$ is bounded, $$\frac1t\operatorname E_x\left[\int_0^tZ_s\:{\rm d}s\right]\xrightarrow{t\to0+}\operatorname E_x[Z_0]\tag5,$$ aren't we able to conclude by replacing $Z$ with $Z^n:=\min(Z,n)$? The monotone convergence theorem should yield $$\operatorname E_x\left[\int_0^tZ^n_s\:{\rm d}s\right]\xrightarrow{n\to\infty}\operatorname E_x\left[\int_0^tZ_s\:{\rm d}s\right]\tag6$$ and $$\operatorname E_x[Z^n_0]\xrightarrow{n\to\infty}\operatorname E_x[Z_0]\tag7.$$ Or am I missing something? Jul 6 at 5:47
• @ChristopheLeuridan If not, it would be interesting if we can drop the continuity assumption by approximating $c$ in a suitable way ... Do you think that's possible? Jul 6 at 7:14
• @ChristopheLeuridan I'd still be interested in your opinion. Jul 16 at 11:15

I don't have enough reputation to comment, this is just an idea. If you assume that$$^1$$

1. $$E$$ is LCCB,
2. $$X$$ is càdlàg,
3. $$X$$ is strong Markov, and
4. $$X$$ is quasi-left continuous,

then for any $$t$$ the set $$A = \{X_s(\omega) : 0\le s \le t < \infty\}$$ is almost surely bounded.$$^2$$ So in that case local boundedness of $$c$$ would be enough to apply dominated convergence, because you could contain $$A$$ in a compact set on which $$c$$ is bounded.

1: Along with your assumptions, these ensure that $$X$$ is a Hunt process, but not necessarily a Feller process.

2: Proposition I-9.3 in Blumenthal and Getoor's book.

• Thank you for your input. I've checked Blumental and Getoor, but unless I'm missing something the bound for the set $A$ does depend on $\omega$. To be precise, if I understand the claim correctly, they say that for each $t\ge0$, there is a null set $N$ such that $\{X_s(\omega):s\in[0,t]\}$ is bounded for all $\omega\not\in N$. So, the bound should depend on $\omega$ and hence this is not enough to apply the DCT. Jul 14 at 19:44
• Please see mathoverflow.net/q/426699/91890. Jul 16 at 11:05
• True it doesn't take you all the way there, but I think it could be used to show that $\mathbb P_x(X_s\in U \text{ for all }0\le s < t)\to 1$ as $t\to0$ for all open $U$ containing $x$. Jul 17 at 14:06
• Don't you think that we need to assume continuity of $c$? Actually, the only thing we need is continuity of $c\circ X$ at $0$, but I don't see any assumption other than continuity of $c$ which would ensure that. Morever, what do you think about the attempt I've described in the comments below the question? Jul 18 at 14:17
• How would you conclude from $\mathbb P_x(X_s\in U \text{ for all }0\le s < t)\to 1$ for each open neighborhood $U$ of $x$? Please see math.stackexchange.com/q/4495504/47771. Jul 18 at 19:26