The formula you give is wrong. It should read $ \{\omega\,:\,g(\omega)\leq r\} = \bigcap_{n=1}^{\infty} \{\omega\,:\,f_{n}(\omega) \leq r\}. $ To prove this, note: If $A \subset B$ and $A \supset B$ then $A = B$. Equivalently: if for all $a \in A$ we have $a \in B$ and if for all $b \in B$ we have $b \in A$ then it follows that $A = B$.
Note that $\omega \in \bigcap_{n=1}^{\infty} \{\omega\,:\,f_{n}(\omega) \leq r\}$ means that for all $n$ we have $f_{n}(\omega) \leq r$, so $g(\omega) = \sup_{n} f_{n}(\omega) \leq r$ as well. This shows $\supset$ above. Conversely, if $g(\omega) \leq r$ then for all $n$ we have $f_{n}(\omega) \leq g(\omega) \leq r$ by the definition of the supremum, so $\omega \in \bigcap_{n=1}^{\infty} \{\omega \,:\, f_{n}(\omega) \leq r\}$ and this shows $\subset$ above.