A family of sets $C$(or say a set of sets) is a monotone class if it is closed under countable monotone limit that is if $A_n ∈ C$, $n \geq 1$, $A_n \uparrow A$(or $A_n \downarrow A$) then $A∈C$ where $A_n \uparrow A$ denotes Set Monotone Increase that is {$A_n$, $n \geq 1$} is a set sequence with $A_n \subset A_{n+1}$ and existence of $limit_n A_n$.
Well, this is the definition of a family of sets being a monotone class. Can I use it from an inverse view that suppose arbitrary $A∈C$(we have known $C$ is a monotone class) then since $C$ is a monotone class, there does exist a monotone set sequence {$A_n$} $\subset C$ such that {$A_n$} $\uparrow$ $A$? I'm not sure whether it is legel to suppose such "{$A_n$} does exist"? I need to know coz I'm proving something and got stuck here.
The sequence $A_n := A$ does the job.
If you are looking for a non-trivial sequence, then the answer is in general "no". Just consider e.g. an arbitrary set $\Omega \neq \emptyset$ and denote by $$C := \sigma(B) = \{\emptyset,B,B^c,\Omega\}$$ the $\sigma$-algebra generated by a set $B \subseteq \Omega$. Then $C$ is a monotone class, but for any sequence $(A_n)_n \subseteq C$ such that $A_n \uparrow A := \Omega$ we have $A_n = \Omega$ for $n$ sufficiently large.