So, I was working on an exercise and I got stuck on trying to prove that a given function was measurable and I took a look at the solutions.
The function at cause is the following: \begin{equation*} f_n:\left[\frac{1}{n},1-\frac{1}{n}\right] \rightarrow \mathbb{R} \hspace{.1cm} \text{ s.t. } \hspace{.1cm} f_n(x) = x^{-1/2} \end{equation*} this being for each $n \in \mathbb{N}$ s.t. $n \geq 1$.
Now, the justification given for the measurability of each $f_n(x)$ is that we have that $f^{-1}(a,\infty)$ is open and thus measurable (for all $a \in \mathbb{R}$).
What I understand/not understand. I recon that an open set surely is a measurable one, but I don't see how we apply that to this case. From my point of view, which might be wrong, $f^{-1}(a,\infty)$ gives us a set of real values for each different $n$. Now, how does this composse a set? How can we look at a "set" of real values/numbers and say it is open?
Thanks for all the help in advance.
As stated in the comments, it is enough to show that the function is continuous. Let $X,Y$ be topological spaces (in your case $\mathbb{R}$ and $\mathbb{R}$).
$f:X\rightarrow Y$ is continuous iff for all open sets $O\subset Y$, $f^{-1}(O)$ is open.
Now, the solution does not claim this is true for all open sets, but just for the sets $(a,\infty)$. However, this is enough:
Let $S$ be a subbase of the topology on $Y$. Then, a function $f:X\rightarrow Y$ is continuous iff for all $O\in S$, $f^{-1}(O)$ is open.
Finally, the collection $\{(a,\infty):a\in\mathbb{R}\}$ is a subbase for the (standard) topology on $\mathbb{R}$.