I'm working on the following problem from Walter Rudin's Real and Complex Analysis, Third Edition.
Prove that if $f$ is a real function on a measurable space $X$ such that $\{x\::\:f(x)\geq r\}$ is measurable for each rational $r$, then $f$ is measurable.
Here's my attempt at a solution:
By definition, $f:X\to\mathbb{R}$ is measurable if the inverse image of an open set in $\mathbb{R}$ is a measurable set in $X$. We recall that a set in $\mathbb{R}$ is open if and only if it is a union of open segments $(a,b)$. Let $\mathscr{O}\subset\mathbb{R}$ be open. Then in particular, we can write $\mathscr{O}$ as the union of a countable number of segments $\alpha_{n}=(q_{n}-r_{n},q_{n}+r_{n})$, where $q_{n}$ and $r_{n}$ are rational numbers for all $n\in\mathbb{Z}^{+}$. We note that $f^{-1}(\mathscr{O})=f^{-1}(\bigcup_{n=1}^{\infty}\alpha_{n})=\bigcup_{n=1}^{\infty}f^{-1}(\alpha_{n})$. Now, for each $\alpha_{n}$, we have $$\begin{split}(f^{-1}(\alpha_{n}))^{c}&=\{x\::\:f(x)\geq q_{n}+r_{n}\}\cup\{x\::\:f(x)\leq q_{n}-r_{n}\}\\ &=\{x\::\:f(x)\geq q_{n}+r_{n}\}\cup\{x\::\:f(x)\geq r_{n}-q_{n}\}.\end{split}$$ By assumption, the sets $\{x\::\:f(x)\geq q_{n}+r_{n}\}$ and $\{x\::\:f(x)\geq r_{n}-q_{n}\}$ are measurable. Since the finite union of measurable sets is measurable, this shows that $(f^{-1}(\alpha_{n}))^{c}$ is measurable for each $n$. But this implies that $f^{-1}(\alpha_{n})=((f^{-1}(\alpha_{n}))^{c})^{c}$ is measurable for each $n$, and, since the countable union of measurable sets is measurable, we see that $\bigcup_{n=1}^{\infty}f^{-1}(\alpha_{n})$ is measurable. Therefore, $f^{-1}(\mathscr{O})$ is measurable, and we conclude that $f:X\to\mathbb{R}$ is a measurable function.
Does this argument look okay? Thanks in advance for any help!