I cannot understand this proof.
Be $(X, \mathcal{F})$ a measurable space. If $f_n : X \to \bar{\mathbb{R}}$, with $n\in\mathbb{N}$ are measurable, then $\sup f_n$ is measurable.
Proof
$\forall a\in\mathbb{R}$ we have $\{x\in X \;:\; \sup_n f_n > a\} \equiv \bigcup_n \{x\in X \;:\; f_n > a\}$ and this belongs to $\mathcal{F}$.
The inclusion $\supset$ is trivial: $a < f_n(x) \leq \sup_n f_n(x)$.
The inclusion $\subset$ comes by contradiction: if $f_n(x) \leq a$ then also $\sup_n f_n(x) \leq a$. This concludes.
- My question is: how the second inclusion concludes the proof? In what way can I say tha then sup is measurable?
I don't really understand this proof, can someone please explain me this better?
To prove that a function $g$ is Borel-measurable, it suffices to show that the sets $\{x\in X\;:\; g(x)>a\}$, $a\in \mathbb R$, are all measurable. That is, it is enough to consider the preimage of the sets $(a,+\infty]$ through the function $g$.
The proof is trying to show this for $g=\sup_n f_n$. How? They rewrite those sets as union of sets that are already known to be measurable, using the fact that the functions $f_n$ are measurable.