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I'm reading about Lévy Processes.

As a definition of a recurrence and transient Lévy process we have:

Def. For a Lévy process $\{X_{t}\}_{t\geq 0}$ in $\mathbb{R}^{d}$ is called recurrent if $\liminf_{t\rightarrow\infty}|X_{t}|=0\space a.s.$

It is called transient if $\displaystyle\lim_{t\rightarrow\infty}|X_{t}|=\infty\space a.s.$

Then the book says that events are measurables because $$\{\liminf_{t\rightarrow\infty}|X_{t}|=0\}=\bigcap_{k=1}^{\infty}\bigcap_{n=1}^{\infty}\bigcup_{t\in\mathbb{Q}\cap(n,\infty)}\{|X_{t}|<1/k\}$$ and $$\{\displaystyle\lim_{t\rightarrow\infty}|X_{t}|=\infty\}=\bigcap_{k=1}^{\infty}\bigcup_{n=1}^{\infty}\bigcap_{t\in\mathbb{Q}\cap(n,\infty)}\{|X_{t}|>k\}.$$

I'm trying to prove these equalities but I don't get the compatibility between the definition of limit and infierior limit with these expressions.

Any kind of help is thanked in advanced.

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Let us prove the first equality under the assumption that the process is right-continuous. The other properties about Levy process (e.g., independent increment) are irrelevant. To facilitate typing, we consider the case $d=1$ only (the general case is the same).

Let $X:[0,\infty)\times\Omega\rightarrow\mathbb{R}$ be a map such that for each $\omega\in\Omega$, $X(\cdot,\omega)$ is a right-continuous function. We assert that $$ \{\omega\in\Omega\mid\lim\inf_{t\rightarrow\infty}|X(t,\omega)|=0\}=\bigcap_{k=1}^{\infty}\bigcap_{n=1}^{\infty}\bigcup_{t\in\mathbb{Q}\cap(n,\infty)}\{\omega\in\Omega\mid|X(t,\omega)|<\frac{1}{k}\}. $$ Proof: Let $\omega\in LHS$. Let $k,n\in\mathbb{N}$ be arbitrary. Since $\lim_{s\rightarrow\infty}\inf_{t\geq s}|X(t,\omega)|=0$, there exists $s_{0}$ such that $\inf_{t\geq s}|X(t,\omega)|<\frac{1}{k}$ whenever $s>s_{0}$. Choose $n_{0}\in\mathbb{N}$ such that $n_{0}>\max(n,s_{0})$. Then $\inf_{t\geq n_{0}}|X(t,\omega)|<\frac{1}{k}$. Now $\frac{1}{k}$ is not a lower bound of the set $\{|X(t,\omega)|\mid t\geq n_{0}\}$, so there exists $t_{0}\geq n_{0}$ such that $|X(t_{0},\omega)|<\frac{1}{k}$. By right-continuity of the function $X(\cdot,\omega)$ and density of $\mathbb{Q}$, we can choose $t_{1}\in\mathbb{Q}\cap(t_{0},\infty)$ such that $|X(t_{1},\omega)|<\frac{1}{k}$. Clearly $t_{1}\in\mathbb{Q}\cap(n,\infty)$, so $\omega\in RHS$.

Conversely, let $\omega\in RHS$. We construct a strictly increasing sequence $(t_{l})$ such that $t_{l}\rightarrow\infty$ and $|X(t_{l},\omega)|\rightarrow0$. It will follow that $\omega\in LHS$. Put $k=n=1$, choose $t_{1}\in\mathbb{Q}\cap(1,\infty)$ such that $|X(t_{1},\omega)|<1$. Suppose that $t_{1},t_{2},\ldots,t_{l}$ have been chosen. Choose $n\in\mathbb{N}$ such that $n>t_{l}+1$. Now $\omega\in\bigcup_{t\in\mathbb{Q}\cap(n,\infty)}\{\omega\mid|X(t,\omega)|<\frac{1}{l+1}\}$, so there exists $t_{l+1}\in\mathbb{Q}\cap(n,\infty)$ such that $|X(t_{l+1},\omega)|<\frac{1}{l+1}$. By recursion theorem, we obtain a sequence $(t_{l})$ which has property: $t_{l+1}>t_{l}+1$ and $|X(t_{l+1},\omega)|<\frac{1}{l+1}$, for all $l\in\mathbb{N}$. Therefore $t_{l}\rightarrow\infty$ and $|X(t_{l},\omega)|\rightarrow0$.

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  • $\begingroup$ So clear! Many thanks! $\endgroup$ – Squird37 May 2 '18 at 16:51
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You need other properties about Levy process. Otherwise, these two identities does not hold. For example, if $X:[0,\infty)\times\Omega\rightarrow \mathbb{R}$ is just a map, the first equality fails to hold. For, we may define $X(t,\omega)= 0$ if $t$ is irrational and $X(t,\omega)= 1$ if $t$ is rational. Then $\{\omega\mid \liminf_{t\rightarrow\infty} |X(t,\omega)|=0\}$ is $\Omega$ but RHS is empty.

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  • $\begingroup$ Will be enough if each set has intersection with the set that gives right-continuity to the process? $\endgroup$ – Squird37 May 2 '18 at 3:20
  • $\begingroup$ Yes, if the process is right-continuous, then the two equalities hold. Some textbooks assume that all Levy processes are cadlag but some do not. You need to specify all the premises at the beginning. $\endgroup$ – Danny Pak-Keung Chan May 2 '18 at 13:22

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