I'm working through Axler's "Measure, Integration, and Real Analysis" and have come across the following theorem:
Suppose $(X,S)$ is a measurable space and $f_1,f_2,\cdots$ is a sequence of $S$-measurable functions from $X$ to $[-\infty, \infty]$. Define $h:X\rightarrow[-\infty, \infty]$ by $h(x)=\sup{\{f_k(x):k\in \mathbf{Z}^+\}}$. Then $h$ is an $S$-measurable function.
He gives a proof using the following equation, which the states the reader should verify ($a \in \mathbf{R}$):
$$h^{-1}((a,\infty])=\bigcup_{k=1}^\infty f_k^{-1}((a,\infty])$$
I'm attempting to verify this. My attempted proof:
Let $x\in \bigcup_{k=1}^\infty f_k^{-1}((a,\infty])$. Then we know that $f_k(x)>a$ for some $k\in\mathbf{Z}^+$. By definition of the supremum, we know that $h(x)\geq f_k(x) \forall k \in \mathbf{Z}^+$. Since $h$ is always greater than or equal to all $f_k$, and there is at least one $f_k>a$, we have that $h(x)>a$, meaning that $x\in h^{-1}((a,\infty])$.
This shows that
$$ \bigcup_{k=1}^\infty f_k^{-1}((a,\infty]) \subseteq h^{-1}((a,\infty]) $$
For the opposite direction, let $x\in h^{-1}((a,\infty])$. Thus, $h(x)>a$. Now, by definition of the supremum, we know that $h(x)\geq f_k(x) \forall k \in \mathbf{Z}^+$. $\cdots$
and that's where I lose the thought process. I'm trying to show that $h^{-1}((a,\infty]) \subset \bigcup_{k=1}^\infty f_k^{-1}((a,\infty])$ thereby proving that the two sets are equal, but I'm missing the last step. If I can figure out how to show that $\exists k\in\mathbf{Z}^+$ such that $f_k(x)>a$, then I would have the proof completed.
Best Answer
It is sufficient to show that there is some $k \ge 1$ such that $f_k(x) > a$. To that end, assume to the contrary that $f_k(x) \le a$ for all $k \ge 1$. But then this implies $\sup_k f_k(x) \le a$ contradicting the fact that $\sup_k f_k(x) = h(x) > a$.