The following question might be too easy for many people and concerns about a very basic doubt, which I am sure I am missing somehow, but I will be glad if someone could help me a little clarifying my confusion.
Let $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $\mathbb{R}$ and let $(S,\Sigma)$ be a measurable space with the set of all $\Sigma$-measurable functions denoted by $m\Sigma$. Then, the proposition 3.2 (b) of the book Probability with Martingales by David Williams states the following
Let $\mathcal{C}\subseteq \mathcal{B}$ and $\sigma(\mathcal{C})=\mathcal{B}$. Then $h^{-1}:\mathcal{C}\to \Sigma$ implies that $h\in m\Sigma$.
The proof proceeds by showing that the class of sets $\mathcal{A}=\{A\in \mathcal{B}:h^{-1}(A)\in \Sigma\}$ is a $\sigma$-algebra containing $\mathcal{C}$ and therefore is equal to $\mathcal{B}$, establishing the desired result. While the proof is very easy, I am confused to see that during the proof the author does not show that $\mathbb{R}\in \mathcal{A}$, or equivalently, $\emptyset\in \mathcal{A}$. I have also seen the same kind of proof in quite a few other books, which means there must be something trivial about this, but I don't understand how to show that $h^{-1}(\mathbb{R})\in \Sigma$ or, equivalently, $h^{-1}(\emptyset)\in \Sigma$ unless $\mathbb{R}\in \mathcal{C}$ or $\emptyset \in \mathcal{C}$. Is it somehow related to the fact that $\sigma(\mathcal{C})=\mathcal{B}$? Can somebody help? Thanks in advance.
$\mathcal{A}$ contains $\emptyset$ because $h^{-1}[\emptyset] = \emptyset \in \Sigma$ ($\sigma$ being a $\sigma$-algebra) and also $h^{-1}[\Bbb R]=S \in \Sigma$ as well. Just basic set theory.