• $(E,\mathcal E)$ 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 $$c_n:=\min(c,n)\in\mathcal E_b\;\;\;\text{for all }n\in\mathbb N.$$

  • $\begingroup$ 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 $s=tu$. $\endgroup$ Commented Jul 5, 2022 at 19:53
  • $\begingroup$ @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 $c$. $\endgroup$
    – 0xbadf00d
    Commented Jul 6, 2022 at 5:07
  • $\begingroup$ @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? $\endgroup$
    – 0xbadf00d
    Commented Jul 6, 2022 at 5:47
  • $\begingroup$ @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? $\endgroup$
    – 0xbadf00d
    Commented Jul 6, 2022 at 7:14
  • $\begingroup$ @ChristopheLeuridan I'd still be interested in your opinion. $\endgroup$
    – 0xbadf00d
    Commented Jul 16, 2022 at 11:15

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – 0xbadf00d
    Commented Jul 14, 2022 at 19:44
  • $\begingroup$ Please see mathoverflow.net/q/426699/91890. $\endgroup$
    – 0xbadf00d
    Commented Jul 16, 2022 at 11:05
  • $\begingroup$ 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 $x$. $\endgroup$
    – user1118
    Commented Jul 17, 2022 at 14:06
  • $\begingroup$ 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? $\endgroup$
    – 0xbadf00d
    Commented Jul 18, 2022 at 14:17
  • $\begingroup$ 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. $\endgroup$
    – 0xbadf00d
    Commented Jul 18, 2022 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.