# Are we able to apply the monotone convergence theorem to show that $\frac1t\text E\left[\int_0^t\min(X_s,n)\:{\rm d}s\right]\to\text E[X_0]$?

Let $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space and $$(X_t)_{t\ge0}$$ be an $$[0,\infty)$$-valued process on $$(\Omega,\mathcal A)$$. Assume $$X:\Omega\times[0,\infty)\to[0,\infty)$$ is $$(\mathcal A\otimes\mathcal B([0,\infty)),\mathcal B([0,\infty)))$$-measurable and $$X(\omega):[0,\infty)\to[0,\infty)$$ is (right-)continuous at $$0$$ for $$\operatorname P$$-almost all $$\omega\in\Omega$$.

Let $$Y_t:=\int_0^tX_s\:{\rm d}s\in[0,\infty]\;\;\;\text{for }t\ge0$$ and $$Z_t:=\operatorname E[Y_t]\;\;\;\text{for }t\ge0.$$

Are we able to show that $$\frac{Z_t}t\xrightarrow{t\to0+}\operatorname E[X_0]\tag1?$$

Note that we can easily show that $$\frac{Y_t(\omega)}t\xrightarrow{t\to0+}X_0(\omega)\tag2$$ for $$\operatorname P$$-almost all $$\omega\in\Omega$$. If $$\sup_{(\omega,\:t)\in[0,\:\infty)}X_t(\omega)<\infty,$$ then we can easily conclude $$(1)$$ by the dominated convergence theorem.

For the general case, let $$X^n:=X\wedge n$$, $$Y^n_t:=\int_0^tX_s^n\:{\rm d}s\;\;\;\text{for }t\ge0$$ and $$Z^n_t:=\operatorname E[Y^n_t]\;\;\;\text{for }t\ge0$$ for $$n\in\mathbb N$$. By the former case, $$\frac{Z^n_t}t\xrightarrow{n\to\infty}\operatorname E[X^n_0]\tag3$$ for all $$n\in\mathbb N$$.

Now the idea is to conclude using the monotone convergence theorem. However, we need to let $$n\to\infty$$ and $$t\to0+$$ simultaneously ... Are we able to show that this is actually possible?

Here is a counter-example to show that $$\lim_{t\rightarrow\infty} E[Y_t/t]$$ may not be $$E[X_0]$$:

Let $$V$$ be any nonnegative random variable with $$E[V]=\infty$$. Define $$X_s=sV$$ for $$s\geq 0$$. Then for each $$\omega \in \Omega$$, we have $$X_s(\omega)=sV(\omega)$$ is continuous at all points $$s\geq 0$$, and $$X_0(\omega)=0$$ for all $$\omega \in \Omega$$. However

$$Y_t = \int_0^t X_s ds = Vt^2/2 \quad \forall t \geq 0$$

So $$Y_t/t = Vt/2$$ and $$E[Y_t/t]=\infty$$ for all $$t>0$$. $$\Box$$

On the other hand, by Fatou's lemma and your equation (2) we get $$E[X_0] = E[\liminf_{t\rightarrow\infty} Y_t/t] \leq \liminf_{t\rightarrow\infty} E[Y_t/t]$$ I suppose we could also conclude this from your equation (3). The counter-example shows the inequality can be strict.

• Thank you for answer. Do you think the situation is different in the following case: $X=f\circ W$, where $E$ is a topological space, $(W_t)_{t\ge0}$ is an $E$-valued càdlàg process on $(\Omega,\mathcal A)$ and $f:E\to[0,\infty)$ is Borel measurable and locally bounded? Commented Jul 18, 2022 at 18:37
• Would my same example with $E=[0, \infty)$, $W_t=tV$, and $f(x)=x$ fit that situation? Commented Jul 18, 2022 at 18:47
• It obviously does ... Sorry. Do you think there is any assumption milder than boundedness on $f$ which would turn the claim to be true? Commented Jul 18, 2022 at 18:50
• What I think could be an interesting additional assumption is that $\operatorname P$ be replaced by a probability measure $\operatorname P_x$ with $W_t=x$ $\operatorname P_x$-almost surely and $\operatorname P_x[X_s\in U\text{ for all }s\in[0,t]]\to1$ as $t\to0+$ for all open neighborhoods $U\subseteq E$ of $x$. Commented Jul 18, 2022 at 18:52
• Please see math.stackexchange.com/q/4495504/47771. Commented Jul 18, 2022 at 19:25