Problem with $\mathbb{P}(\liminf_n(A_n\cup B_n))$

Question

We know that $\mathbb{P}(\liminf_n A_n)=0.3$ and $\mathbb{P}(\limsup_n B_n)=0$. Find $\mathbb{P}(\liminf_n(A_n\cup B_n))$.

My solution: We know that $\liminf_n(A_n\cup B_n)\supset \liminf_nA_n \cup \liminf_n B_n \supset P(\liminf_n A_n)$. Therefore $$\mathbb{P}(\liminf_nA_n)=0.3\leqslant \mathbb{P}(\liminf_n(A_n\cup B_n)).$$ How should I show that $\mathbb{P}(\liminf_n(A_n\cup B_n)) \leqslant 0.3$?

Answer 1

First, $\Bbb P(\liminf_nB^c_n)=1$ because $\Bbb P(\limsup_nB_n)=0$. Therefore $$ \Bbb P(\liminf_n(A_n\cup B_n))=\Bbb P([\liminf_n (A_n\cup B_n)]\cap[\liminf_nB^c_n]). $$ But $$ [\liminf_n (A_n\cup B_n)]\cap[\liminf_nB^c_n]\subset \liminf A_n $$ and the upper bound follows.

Answer 2

We claim the following relation

$$ \liminf_{n\to\infty} (A_n \cup B_n) \subseteq \Big( \liminf_{n\to\infty} A_n \Big) \cup \Big( \limsup_{n\to\infty} B_n \Big). \tag{*} $$

Notice that this immediately yields the inequality $\Bbb{P}(\liminf_n (A_n \cup B_n)) \leq 0.3$. So it suffices to prove $\text{(*)}$.

A key observation is the following identity: for any fixed $N$ and for any sequence of sets $(E_n)$,

$$ \limsup_{n\to\infty} E_n = \bigcap_{n = N}^{\infty} \bigcup_{j=n}^{\infty} E_j \quad \text{and} \quad \liminf_{n\to\infty} E_n = \bigcup_{n = N}^{\infty} \bigcap_{j=n}^{\infty} E_j. $$

Now let us denote $\tilde{B}_n = \cup_{j \geq n} B_j$. Then

\begin{align*} \liminf_{n\to\infty} (A_n \cup B_n) &= \bigcup_{n=N}^{\infty} \bigcap_{j=n}^{\infty} (A_j \cup B_j) \subseteq \bigcup_{n=N}^{\infty} \bigcap_{j=n}^{\infty} (A_j \cup \tilde{B}_N) \\ &= \left( \bigcup_{n=N}^{\infty} \bigcap_{j=n}^{\infty} A_j \right) \cup \tilde{B}_N = \Big( \liminf_{n\to\infty} A_n \Big) \cup \tilde{B}_N \end{align*}

Taking intersection $\cap_{N=1}^{\infty}$ to both sides proves $\text{(*)}$ as desired.