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

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  • $\begingroup$ 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 $s=tu$. $\endgroup$ Jul 5 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
    Jul 6 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
    Jul 6 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
    Jul 6 at 7:14
  • $\begingroup$ @ChristopheLeuridan I'd still be interested in your opinion. $\endgroup$
    – 0xbadf00d
    Jul 16 at 11:15

1 Answer 1

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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.

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  • $\begingroup$ 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
    Jul 14 at 19:44
  • $\begingroup$ Please see mathoverflow.net/q/426699/91890. $\endgroup$
    – 0xbadf00d
    Jul 16 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
    Jul 17 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
    Jul 18 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
    Jul 18 at 19:26

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