Let $\{f_n\}$ be a sequence of measurable function defines on a set $X$. it's simple to prove that $$\{\sup f_n > \alpha\}=\bigcup_{n=1}^\infty \{f_n> \alpha\},\tag1$$ where $\alpha\in \mathbb{R}$. Can I conclude from this that also $$\{\sup(-f_n)>-\alpha\}=\bigcup_{n=1}^\infty\{-f_n>-\alpha\}.$$
In my opinion the answer is yes and the proof is the same as that used to prove (1), but I'm not sure. Could you give me a hand?