Proof of characterization of measurable sets I am not sure if the following characterization works. 
Suppose the $\mathcal{M}$ is a $\sigma$-algebra in $X$. Let $f$ be a map from $X$ to $[-\infty,\infty]$, that is,
$$f:(X,\mathcal{M}) \rightarrow [-\infty,\infty]$$
Then, $f$ is measurable if  $\forall \alpha \in \mathbb{R}$,we have $f^{-1}((\alpha,\infty)) \in \mathcal{M}$. 
Proof:
Assume that $f$ is measurable and let $\alpha$ to be arbitrary. Note that since   $f$ is measurable, we have $f^{-1}((\alpha,\infty)) \in \mathcal{M}$. Since $\alpha$ was arbitrary, then $\forall \alpha \in \mathbb{R}$,we have $f^{-1}((\alpha,\infty)) \in \mathcal{M}$.  Now, assume that $\forall \alpha \in \mathbb{R}$, we have $f^{-1}((\alpha,\infty)) \in \mathcal{M}$. In particular, we have that for every $\alpha$, the subset $(\alpha,\infty) \subset [-\infty,\infty]$ is a borel set, so that the collection $f^{-1}((\alpha,\infty)) \in \mathcal{M}$ and thus $f$ is measurable.In particular the set:
$$\Omega = \bigcup_{\alpha \in \mathbb{R}} \Omega_{\alpha}, \text{ where } \Omega_{\alpha} = \{(\alpha,\infty) \subset [-\infty,\infty]| f^{-1}((\alpha,\infty)) \in \mathcal{M}\},$$
is a $\sigma-$algebra.
Is this proof correct?
 A: In fact the given condition does not imply that $f$ is measurable. We're given that $f:X\to[-\infty,\infty]$, so $f$ is measurable if the inverse image of every Borel subset of $[-\infty,\infty]$ is measurable. This does not follow from the given condition; the suggested characterization is false.
Counterexample: Say $E$ is some non-measurable set, and let $F=X\setminus E$. Let $f=\infty\chi_E-\infty\chi_F$. Then $f^{-1}((a,\infty))=\emptyset$ for every $a\in\Bbb R$. But $\{\infty\}$ is a Borel subset of $[-\infty,\infty]$, and $f^{-1}(\{\infty\})=E$.
A: You must do this.  Let $\mathcal{S} = \{B\in \mathcal{B}(\mathbb{R})| f^{-1}(B)\in \mathcal{M}\}$.  Then it is easy to show $\mathcal{S}$ is a $\sigma$-algebra.  If $f^{1}(-\infty, a)\in \mathcal{M}$ for all $a\in \mathbb{R}$, it is easy to show that the inverse image of any open interval of under $f$ is in $\mathcal{M}$.  Hence, $\mathcal{S}$ is a $\sigma$-algebra containing all open intervals. You must have $\mathcal{B}(\mathbb{R})\subseteq \mathcal{S}$.
