TITLE: Proof of a theorem on measurable functions
QUESTION [6 upvotes]: Suppose $(\Omega, \mathscr{L})$ and $(S, \mathscr{B})$ are measure spaces and that a collection of sets $\mathscr{A}$ generates $\mathscr{B}$, i.e., $\sigma(\mathscr{A}) = \mathscr{B}$. Let $X: \Omega \rightarrow S$. If $X^{-1}(A) \in \mathscr{L} \quad \forall \space A \in \mathscr{A}$, then $X$ is measurable $\mathscr{B}$ (or $X$ is a random variable).

REPLY [8 votes]: The collection of subsets of $S$ that have a preimage under $X$ in $\sigma$-algebra $\mathcal L$ can be shown to be a $\sigma$-algebra. 
This is not really difficult to prove, since preimages are very coöperative in this. 
So if this collection contains $\mathcal A$ as a subcollection then it will also contain $\mathcal B=\mathcal\sigma(\mathcal A)$ as a subcollection.
That means exactly that $X$ is measurable.