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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
  • $X,X^n:\Omega\times[0,\infty)\to\mathbb R$ be stochastic processes on $(\Omega,\mathcal A,\operatorname P)$ such that
    1. $(X^n_t)_{n\in\mathbb N}$ is nondecreasing for all $t\ge0$
    2. $X$ is left-continuous
    3. $X^n$ is nondecreasing (i.e. $t\mapsto X^n_t$ is nondecreasing) for all $n\in\mathbb N$
    4. $X-X^n$ is nondecreasing and $(X-X^n)_0=0$ for all $n\in\mathbb N$

Assume $$\operatorname P\left[\left|X_t-X_t^n\right|>\varepsilon\right]\xrightarrow{n\to\infty}0\;\;\;\text{for all }\varepsilon>0\text{ and }t\ge0\;.\tag1$$

I want to show that $$X^n_t\xrightarrow{n\to\infty}X_t\;\;\;\text{for all }t\ge0\text{ almost surely}\;.\tag2$$

By 1. and $(1)$, we obtain $$X_t^n\xrightarrow{n\to\infty}X_t\;\;\;\text{almost surely for all }t\ge0\tag3$$ (see this question).

So, the crucial point is the selection of a set $\Omega_0\in\mathcal A$ with $\operatorname P[\Omega_0]=1$ such that $$X_t^n(\omega)\xrightarrow{n\to\infty}X_t(\omega)\;\;\;\text{for all }(\omega,t)\in\Omega_0\times[0,\infty)\;.\tag4$$

Let $$\Omega_0:=\bigcup_{t\in[0,\:\infty)\cap\mathbb Q}\left\{\omega\in\Omega:X^n_t(\omega)\xrightarrow{n\to\infty}X_t(\omega)\right\}$$ and $(\omega,t)\in\Omega_0\times[0,\infty)$.

Let $\varepsilon>0$. We need to show that $$|X_t(\omega)-X^n_t(\omega)|<\varepsilon\;\;\;\text{for all }n\ge n_0\tag5$$ for some $n_0\in\mathbb N$.

  • Since $\mathbb Q$ is dense in $\mathbb R$ and $X(\omega)$ is left-continuous, there is some $s\in[0,t]\cap\mathbb Q$ with $$|X_t(\omega)-X_s(\omega)|<\frac\varepsilon2\tag6$$
  • Since $\omega\in\Omega_0$ and $s\in[0,\infty)\cap\mathbb Q$, there is some $n_0\in\mathbb N$ with $$|X_s(\omega)-X^n_s(\omega)|<\frac\varepsilon2\;\;\;\text{for all }n\ge n_0\tag7$$

Now, I think we have no chance to conclude without 3. and 4. (please correct me, if I'm wrong).

By 3. and 4., $$0\le X_t(\omega)-X^n_t(\omega)=\underbrace{\left(X_t(\omega)-X_s(\omega)\right)}_{<\:\frac\varepsilon2}+\underbrace{\left(X_s(\omega)-X^n_s(\omega)\right)}_{<\:\frac\varepsilon2}+\underbrace{\left(X^n_s(\omega)-X^n_t(\omega)\right)}_{\le\:0}<\varepsilon\tag8$$ for all $n\ge n_0$.

Is there any mistake? And is the conclusion even possible without 3. or 4.?

0xbadf00d
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    It's a little confusing: sometimes nondecreasing means nondecreasing in $n$ and sometimes nondecreasing in $t$. If all you have is the $t$ kind of nondecreasing, constant (I mean, constant in $t$) functions seem legit, and then any i.p. does not imply a.s. example gives a counterexample. So it sounds like you need at least 3. or 4. – kimchi lover Sep 07 '17 at 22:25
  • @kimchilover Only in 1. the meaning of nondecreasing is nondecreasing in $n$ (which should be clear, since the $t$ is fixed and the object in question is a sequence). At any other point, nondecreasing has to be understood as a path property. – 0xbadf00d Sep 07 '17 at 22:56

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