continuity of probability distribution

Hello,

For the last inequality $P(\liminf_n A_n)\leq \liminf_n P(A_n)$ :

Consider $B_N=\bigcap_{n=N}^\infty A_n$
The sequence $(B_N)$ is increasing.
So $P(\lim_N B_N)=\lim_N P(B_N)$ (basic property of a measure/probability)

But $\lim_N B_N=\liminf_n A_n$

So we have $P(\liminf_n A_n)=\lim_N P(B_N)$

But $\forall N,B_N \subseteq A_N$ . Thus $P(B_N)\leq P(A_n)$

$\Rightarrow \lim_N P(B_N)=\liminf_N P(B_N) \leq \liminf_n P(A_n)$

Finally, $P(\liminf_n A_n)\leq \liminf P(A_n)$

(note : $\liminf \equiv \underline{\lim}$ )

Does this help ?

For the first inequality, consider taking the compliment of the last inequality.