Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $a$ and $b$ be two rationals (i.e. $a$,$b$ $\in \mathbb{Q}$) such that $a<b$ and $(X_n)$ be a sequence of random variables defined on the above-defined measurable space. Set: \begin{equation} \Lambda_{a,b}=\{\limsup\limits_{n\rightarrow\infty}X_n\geq b; \liminf\limits_{n\rightarrow\infty}X_n \leq a\} \end{equation} \begin{equation} \Lambda=\bigcup\limits_{a<b} \Lambda_{a,b} \end{equation}
Therefore, I "read" $\Lambda$ as follows:
there exists at least a pair of rationals $a$, $b$ ($a<b$) such that $\limsup\limits_{n\rightarrow\infty}X_n\geq b; \liminf\limits_{n\rightarrow\infty}X_n \leq a$, that is
\begin{equation} \Lambda=\{\limsup\limits_{n\rightarrow\infty}X_n> \liminf\limits_{n\rightarrow\infty}X_n\} \end{equation}
I am given that $\mathbb{P}(\Lambda_{a,b})=0$ $\forall a, b \in \mathbb{Q}$ such that $a<b$. And, at this point, I know that one can state that
"$\mathbb{P}(\Lambda)=0$, since all rational pairs are countable"
I interpret the above statement in the following way, but I am not sure whether it is correct or not.
Since:
- $\mathbb{P}(\Lambda_{a,b})=0$ $\forall a,b \in \mathbb{Q}$ such that $a<b$;
- $\bigcup\limits_{a<b}\Lambda_{a,b}$ is a countable union, since rationals are countable by definition;
by countable subadditivity property of probability measure $\mathbb{P}$, it follows that: \begin{equation} \mathbb{P}(\Lambda)=\mathbb{P}\big(\bigcup\limits_{a<b} \Lambda_{a,b}\big)\leq \sum\limits_{a<b}\mathbb{P}\big(\Lambda_{a,b}\big)=0 \end{equation} where $\sum\limits_{a<b}\mathbb{P}\big(\Lambda_{a,b}\big)=0$ follows from the fact that I am given that $\mathbb{P}(\Lambda_{a,b})=0$ $\forall a, b \in \mathbb{Q}$ such that $a<b$.
Hence, since probability lies by definition between $0$ and $1$, $\mathbb{P}(\Lambda)\leq 0$ "means" that: \begin{equation} \mathbb{P}(\Lambda)=0 \end{equation}
Is my reasoning correct?