would you help me how I can prove this sentence? please.
Hello,
For the last inequality $\displaystyle P(\liminf_n A_n)\leq \liminf_n P(A_n)$ :
Consider $\displaystyle B_N=\bigcap_{n=N}^\infty A_n$
The sequence $\displaystyle (B_N)$ is increasing.
So $\displaystyle P(\lim_N B_N)=\lim_N P(B_N)$ (basic property of a measure/probability)
But $\displaystyle \lim_N B_N=\liminf_n A_n$
So we have $\displaystyle P(\liminf_n A_n)=\lim_N P(B_N)$
But $\displaystyle \forall N,B_N \subseteq A_N$. Thus $\displaystyle P(B_N)\leq P(A_n)$
$\displaystyle \Rightarrow \lim_N P(B_N)=\liminf_N P(B_N) \leq \liminf_n P(A_n)$
Finally, $\displaystyle P(\liminf_n A_n)\leq \liminf P(A_n)$
(note : $\displaystyle \liminf \equiv \underline{\lim}$)
Does this help ?
For the first inequality, consider taking the compliment of the last inequality.