I am trying to show that if a sequence of measurable functions converges uniformly to $f$, then $f$ is measurable. To prove this, I will show that $\sup_{n\ge1}\inf_{k\ge n}f_k$ is measurable since $\sup_{n\ge1}\inf_{k\ge n}f_k = f$. Let $\sup_{n\ge1}\inf_{k\ge n}f_k= g$. Then, we have to show that $\{x| g(x) > \alpha\}$ is measurable. By definition of supremum, for $\varepsilon>0$, there exists $n \in N$ such that $$\inf_{k\ge n}f_k>\sup_{n\ge1}\inf_{k\ge n}f_k-\varepsilon>\alpha-\varepsilon.$$Since $\varepsilon$ is arbitrary, we can say that $\forall j\ge1,$ there exists $n\in N $ such that $$\inf_{k\ge n}f_k>\alpha-1/j.$$ This implies that $\forall k \ge 1, \forall j\ge1, \exists n\in N$ s.t $f_k> \alpha -1/j$. Therefore, $$\bigcup_{k,j \ge1} \bigcap_{n\ge1}\{x|f_k(x)>\alpha-1/j\}=\{x| g(x) > \alpha\}.$$
I have two questions in this proof. First, why $\forall k \ge 1$? I don't know how to interpret this step. The second question is why $\forall$ is expressed as union and $\exists$ is expressed as intersection?
Could you elaborate on this?