Proof of Generator Criterion for Measurability

Write $\mathcal{H}$ for the family of all subsets $B$ of $Y$ with $f^{-1}(B)\in\mathcal{F}$.

Claim 1. Preimages satisfy three identities, each verified by testing membership of a point $x\in X$ through the location of $f(x)$: first, $f^{-1}(Y)=X$, since $f(x)\in Y$ always; second, $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$, since $f(x)\notin B$ is the negation of $f(x)\in B$; third, for any sequence $(B_m)_{m\in\mathbb{N}}$ of subsets of $Y$,

since $f(x)$ lies in the union exactly when $f(x)\in B_m$ for some $m$.

Now $Y\in\mathcal{H}$ because $f^{-1}(Y)=X\in\mathcal{F}$. If $B\in\mathcal{H}$ then $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)\in\mathcal{F}$, since $\mathcal{F}$ is closed under complements, so $Y\setminus B\in\mathcal{H}$. If $(B_m)_{m\in\mathbb{N}}$ is a sequence in $\mathcal{H}$ then the preimage of $\bigcup_m B_m$ is the union of the sets $f^{-1}(B_m)\in\mathcal{F}$, which lies in $\mathcal{F}$ because $\mathcal{F}$ is closed under countable unions, so $\bigcup_m B_m\in\mathcal{H}$. These are the three defining properties in Sigma-Algebra and Measurable Space, so $\mathcal{H}$ is a $\sigma$-algebra on $Y$.

Claim 2. The hypothesis says exactly that $\mathcal{C}\subseteq\mathcal{H}$, and $\mathcal{H}$ is a $\sigma$-algebra on $Y$ by claim 1. Since $\mathcal{G}=\sigma(\mathcal{C})$ is the smallest $\sigma$-algebra on $Y$ containing $\mathcal{C}$ by Generated Sigma-Algebra, it follows that $\mathcal{G}\subseteq\mathcal{H}$, that is, $f^{-1}(B)\in\mathcal{F}$ for every $B\in\mathcal{G}$. This is measurability of $f$ with respect to $\mathcal{F}$ and $\mathcal{G}$.