# Measurability of inferior limit and limit of a Lévy Process

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.

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 $|X(t_{l},\omega)|\rightarrow0$.
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.