Lemma: Let ($\Omega_1,\mathfrak{F}_1$) and ($\Omega_2,\mathfrak{F}_2$) be measurable spaces. Suppose that $\mathfrak{F}_2=\sigma(\mathfrak{A})$ (i.e. the $\sigma$-algebra generated by a set of subsets of $\Omega_2$ called $\mathfrak{A}$). Then $f:\Omega_1 \to \Omega_2$ is a measurable function iff $f^{-1}(A)\in\mathfrak{F}_1, \forall A\in \mathfrak{A}$.
The forward proof is elementary and simple, but there is one aspect of the backwards proof that alludes me. In the proof, we consider a set $\mathfrak{G}:=$ { ${ A\subset\Omega_2 : f^{-1}(A)\in \mathfrak{F}_1}$ }. Once you accept that $\mathfrak{G}$ is a $\sigma$-algebra, the backwards proof follows by $\mathfrak{A}\subset \mathfrak{G} \implies \sigma (\mathfrak{A})\subset \sigma (\mathfrak{G}) = \mathfrak{G}$. As $\mathfrak{F}_2=\sigma(\mathfrak{A}) \implies \mathfrak{F}_2 \subset \mathfrak{G}$. This satisfies the definition of measurable functions.
However, the reason $\mathfrak{G}$ is a $\sigma$-algebra alludes me. Any help would be appreciated. Thank you.
You should verify that the preimage preserves unions and complements. Then for $\{A_n \}_{n=0}^\infty\subseteq \mathfrak{G}$
$$f^{-1}\big( \cup A_n \big)=\cup f^{-1}(A_n)\in \mathfrak{F}_1 \quad \text{and} \quad f^{-1}(A_0^c)=\big( f^{-1}(A_0) \big)^c \in \mathfrak{F}_1,$$
since $\mathfrak{F}_1$ is a $\sigma$-algebra.