$h:S\to \mathbb R$. $\sigma(\mathcal C)$ is the $\sigma$-algebra generated by $\mathcal C$. $m\Sigma$ is the class of $\Sigma$-measurable functions on $S$.
If $\mathcal C \subset \mathcal B$ and $\sigma(\mathcal C) = \mathcal B$, then $h^{-1}: \mathcal C \to \Sigma \implies h \in m\Sigma$.
Proof: Let $\mathcal E$ be the class of elements $B$ in $\mathcal B$ such that $h^{-1}(B) \in \Sigma$. Then $\mathcal E$ is a $\sigma$-algebra since $h^{-1}$ preserves all set operations, and, by hypothesis, $\mathcal E \supseteq \mathcal C.$
To show that $h\in m\Sigma$ if the hypotheses are true, we need to show that for any $B\in \mathcal B$, $h^{-1}(B) \in \Sigma$. How $\mathcal E$ being a $\sigma$-algebra and $\mathcal E \supseteq \mathcal C$ imply that $h\in m\Sigma$?
$\mathcal{E}$ is defined as
$$\mathcal{E}:= \left\{B \in \mathcal{B}\mid h^{-1}[B] \in \Sigma\,\right\}$$
If we know that $\mathcal{C} \subseteq \mathcal{E}$ (the assumption) then $$\sigma(\mathcal{C}) \subseteq \mathcal{E}$$
because $\mathcal{E}$ is a $\sigma$-algebra that contains $\mathcal{C}$ and $\sigma(\mathcal{C})$ is by definition the smallest such $\sigma$-algebra. Now we use that $\sigma(\mathcal{C}) = \mathcal{B}$ (other assumption) and so $\mathcal{B} \subseteq \mathcal{E}$ or by the definition
$$\forall B \in \mathcal{B}: B \in \mathcal{E} \implies \forall B \in \mathcal{B}: h^{-1}[B] \in \Sigma$$
as required for $h \in m\Sigma$.