Consider the measure space ($\Omega, \mathcal{A}$), $A \in \mathcal{A}$. Let $f_n: A \rightarrow [-\infty; \infty]$ be a sequence of measurable functions. My script claims $$\{x \in A: \inf_n \sup_{k \geq n} f_k(x) > t\} = \bigcap_{n \in \mathbb{N}} \bigcup_{k \geq n} \{x \in A: f_k(x) > t\}$$
I struggle to show the strict inequality for "$\supset$". My attempt was as follows: Let $n \in \mathbb{N}$ be arbitrary. Then there exists $k_0 \geq n$ such that $$t < f_{k_0}(x) \leq \sup_{k \geq n} f_k(x)$$ Hence $t$ is a lower bound of the set $$\{\sup_{k \geq n} f_k(x) \mid n \in \mathbb{N}\}$$ As the infium is the greatest lower bound it must hold $$t \leq \inf_{n \in \mathbb{N}}\sup_{k \geq n} f_k(x)$$ I don't see why equality would be strict.
You cannot prove this. Suppose $f_k(x)=t+\frac 1 k$ for all $x$. Then the right side is $\Omega$ and the left side is emptyset!.