• $E$ be a topological space with $\{x\}\in\mathcal B(E)$ for all $x\in E$ and $$\mathcal N(x):=\{N\subseteq E:N\text{ is an open neighborhood of }x\};$$
  • $(\Omega,\mathcal A)$ be a measurable space;
  • $(X_t)_{t\ge0}$ be an $E$-valued right-continuous process on $(\Omega,\mathcal A)$;
  • $\operatorname P_x$ be a probability measure on $(\Omega,\mathcal A)$ with $$\operatorname P_x[X_0=x]=1\tag1$$ and $$\operatorname P_x[X_s\in N\text{ for all }s\in[0,t)]\xrightarrow{t\to0+}1\;\;\;\text{for all }N\in\mathcal N(x)\tag2$$ for $x\in E$;
  • $c:E\to[0,\infty)$ be Borel measurable and locally bounded.

Let $$Y_t:=\int_0^tc(X_s)\:{\rm d}s\in[0,\infty]\;\;\;\text{for }t\ge0$$ and $$Z_t:=\operatorname E_x[Y_t]\;\;\;\text{for }t\ge0$$ for some fixed $x\in E$.

Are we able to show that $$\frac{Z_t}t\xrightarrow{t\to0+}\operatorname E[X_0]\tag3?$$

Note that we can easily show that $$\frac{Y_t(\omega)}t\xrightarrow{t\to0+}X_0(\omega)\tag4$$ for $\operatorname P$-almost all $\omega\in\Omega$.

Since $c$ is locally bounded, there is a $N\in\mathcal N(x)$ with $$c_N:=\sup_Nc\in[0,\infty).$$ Now, the idea is to apply the dominated convergence theorem. In order to do so, we need to find an $\operatorname P_x$-integrable for $\frac{Y_t}t$ for all $t\in(0,\delta)$ for some $\delta>0$.

Let $\varepsilon>0$. By $(2)$, $$\operatorname P_x[X_s\in N\text{ for all }s\in[0,t)]\ge1-\varepsilon\tag5\;\;\;\text{for all }t\in[0,\delta)$$ for some $\delta>0$. Now, we may clearly write $$\frac{Y_t}t\le1_N(X_s)c_N+\frac1t\int_0^t1_{N^c}(X_s)c(X_s)\:{\rm d}s\tag6,$$ but is this helpful?

EDIT: Let $$\tau:=\inf\underbrace{\{t\ge0:X_t\in N^c\}}_{=:\:I}.$$ Since $N^c$ is closed, it holds for all $\omega\in\Omega$ either $\tau(\omega)\in I(\omega)$ or $I(\omega)=\emptyset$ and hence $\tau(\omega)=\infty$. We now see that $$\frac1tY_t\le c_N\;\;\;\text{for all }t\in(0,\tau)\tag7,$$ but is this enough?

EDIT 2: Since $x\not\in N^c$, we see that $\tau(\omega)>0$ for all $\omega\in\{X_0=x\}$ and hence for $\operatorname P_x$-almost all $\omega\in\Omega$.

  • $\begingroup$ I'm not sure what the role of $c$ is, or what $P_x$ really means, but perhaps modifying my example from before to $X_s = X_0 + sV$ for $V$ a nonnegative random variable with $E[V]=\infty$ would be a counter-example. The other answer was math.stackexchange.com/questions/4495450/… $\endgroup$
    – Michael
    Commented Jul 19, 2022 at 1:45
  • $\begingroup$ @Michael $\operatorname P_x$ is just a probability measure und which the set $\{X_0=x\}$ has probability $1$. $\endgroup$
    – 0xbadf00d
    Commented Jul 19, 2022 at 5:38
  • $\begingroup$ So to be clear, $E_x[X_0] = x$? $\endgroup$ Commented Jul 19, 2022 at 15:50
  • $\begingroup$ @user6247850 Since $E$ is a topological space and hence I don't know how you intend to define $\operatorname E_x[X_0]$, no. But if $E$ is a Banach space, yes. $\endgroup$
    – 0xbadf00d
    Commented Jul 19, 2022 at 17:13
  • 1
    $\begingroup$ Taking $c$ as the identity function and assuming $X_0$ and $V$ are independent random variables, and $E[V]=\infty$, doesn't $X_s = X_0 + sV$ act as a counter-example? For $\epsilon>0$, $x\in \mathbb{R}$, and $s>0$ we get $P[X_s \in (X_0-\epsilon, X_0+\epsilon)|X_0=x] = P[|V|<\epsilon/s]\rightarrow 1$ as $s\rightarrow 0^+$. And $Y_t/t= X_0+ Vt/2$ for all $t>0$, $E[Y_t/t|X_0=x]=\infty$ for all $t>0$ and $x\in \mathbb{R}$. $\endgroup$
    – Michael
    Commented Jul 19, 2022 at 22:45


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