If $Y=[-\infty, +\infty]$ and $f^{-1}((\alpha,\infty]) \in M$ for every real $\alpha$, then $f$ is measurable.
Extended real line, $Y=\Bbb{R}\cup \{-\infty, \infty\}$, is simply ordered set with greatest and smallest element. So $\mathcal{T}_Y$ is generated by $\mathcal{B}_Y=\{ (\alpha, \beta)|\alpha \lt \beta$ and $\alpha, \beta \in Y\}$ $\cup$ $\{[-\infty, p)|p \in Y\setminus\{-\infty\}\}$ $\cup$ $\{(q, \infty]| q \in Y\setminus \{\infty\}\}$ basis. $\mathcal{B}_Y$ is not countable. I known $\Bbb{R}$ is second countable. How to show $Y=\Bbb{R} \cup \{-\infty, \infty\}$ is second countable? I think $\mathcal{B}_Y^{\prime}= \{ (\alpha, \beta)|\alpha \lt \beta$ and $\alpha, \beta \in \Bbb{Q}\cup \{-\infty,\infty\}\}$ $\cup$ $\{[-\infty, p)|p \in \Bbb{Q}\cup \{\infty\}\}$ $\cup$ $\{(q, \infty]| q \in \Bbb{Q}\cup \{-\infty\}\}$ is countable basis of $\mathcal{T}_Y$. Am I right?
We actually don’t need to find countable basis. We are just doing it to show $Y$ is second countable. If $X$ is second countable, then $\forall$ basis $\mathcal{C}$ of $\mathcal{T}_X$, $\exists \mathcal{B}$ such that $\mathcal{B}\subseteq \mathcal{C}$ and $\mathcal{B}$ is countable basis of $\mathcal{T}_X$. This is precisely exercise 2 section 30 of Munkres’ topology. So we can write $V\in \mathcal{T}_X$ as $V=\bigcup_{i\in I}B_i$ ; $B_i\in \mathcal{B}$, $\forall i\in I$. So once we know $Y$ is second countable, we don’t have to explicitly find an countable basis of $\mathcal{T}_Y$. I didn’t observe this fact in Theorem 1.8 of Papa Rudin post.
Conclusion: Suppose $(Y,\mathcal{T}_Y)$ is second countable and $\mathcal{B}_Y$ is basis of $\mathcal{T}_Y$. $f:X\to Y$ is measurable $\iff$ $f^{-1}(V)\in M$, $\forall V\in \mathcal{B}_Y$.
One can prove this theorem without taking about $\Omega=\{E\subseteq Y| f^{-1}(E)\in M\}$ and existence of $\{\alpha_n\}$ sequence such that $\alpha_n \to \alpha$. Claim: $f^{-1}([-\infty, \alpha))\in M$, $\forall \alpha \in \Bbb{R} \implies f^{-1}([-\infty, \alpha))\in M$, $\forall \alpha \in \Bbb{R}$. Proof: let $\alpha \in \Bbb{R}$. By elementary set theory, $f^{-1}([-\infty, \alpha))$$=f^{-1}(\bigcup_{n\in \Bbb{N}}[-\infty,\alpha -\frac{1}{n}])$$=\bigcup_{n\in \Bbb{N}}f^{-1}([-\infty,\alpha -\frac{1}{n}])$$=\bigcup_{n\in \Bbb{N}} f^{-1}(\Bbb{R}-(\alpha -\frac{1}{n}, \infty])$. Since $f^{-1}((\alpha -\frac{1}{n}, \infty])\in M$, we have $X-f^{-1}((\alpha -\frac{1}{n}, \infty])=f^{-1}(\Bbb{R}-(\alpha -\frac{1}{n},\infty])\in M$, $\forall n\in \Bbb{N}$. By definition of $\sigma$-algebra, $f^{-1}([-\infty,\alpha))= \bigcup_{n\in \Bbb{N}} f^{-1}(\Bbb{R}-(\alpha -\frac{1}{n}, \infty])\in M$. Our desired result. Is this proof better version of Rudin’s proof?