Let $A_n$ be sequence of events, My book has the following:
$$P(\lim\sup\ A_n) = P(A_n\ {\rm occurs\ for\ infinity\ many\ } n)$$
Why is this equality correct?
I know that: $\lim\sup\ A_n$, is equal to (not formal): All results that belong to infinite events in the sequence $A_n$.
How is this the same as: $A_n$ occurs for infinity many $n$? and what's the meaning of this at all? maybe it's a translation problem (although my professor is native speaker)...
Your professor's formulation means: $$x\in\limsup A_n\iff\exists I\text{ infinite }\subset\Bbb N,\quad x\in\bigcap_{n\in I}A_n.$$ This equivalence is correct because:
I cannot tell you why it is "the same as" your own formulation ("All results that belong to infinite events in the sequence $A_n$") because that one is less clear to me than the usual sentence "$A_n$ occurs for infinity many $n$" for which you asked for an explanation.
The definition of $\limsup A_n$ for a sequence of sets $\{ A_n \}$ is $$ \limsup A_n = \bigcap _{n = 1} ^\infty \bigcup _{m \geq n} A_m. $$
Under this formulation, it is clear that $x \in \limsup A_n$ if and only if for every $n \in \mathbb{N}$, there is a positive number $m \geq n$ which $x \in A_m$. Inuitively speaking, if we now pick $m = n$ assuming they are fixed in the last step, then there is $m_2 \geq m + 1$ which $x \in A_{m_2}$. This tells you that $x$ must be in infinitely many $A_m$, since this process can be continued indefinitely.
On the other hand, if $x$ is not in $\limsup A_n$, then by De Morgan's law $$ x \in (\limsup A_n)^c = \bigcup _{n = 1} ^\infty \bigcap _{m \geq n} A_m^c. $$ This is the set which there is $n$, which $x$ is not in $A_m$ for all $m \geq n$, meaning $x$ stops belonging to $A_n$ after a finite number of them. This shows that if $x \not \in \limsup A_n$ then $x$ is not in infinitely many $A_n$'s.
I think this problem arises mainly because the book does not bring up the set formulation of $\limsup A_n$, and simply define $\limsup A_n$ as the set which $x$ belongs to infinitely many $A_n$'s.
