TITLE: Proving measurability of a function only by checking generating sets
QUESTION [7 upvotes]: Theorem:
Suppose that $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces and $\mathcal{B}$ = $\sigma(\mathcal{G})$ is generated by a family $\mathcal{G}\subset\mathcal{P}(Y)$. Then $f : X \rightarrow Y$ is measurable if and only if
$f^{-1}(G)\in \mathcal{A}$ for every $G \in \mathcal{G}$.
The proof goes as follows:
Set operations are natural under pull-backs, meaning that
$f^{-1}(Y\setminus B) = X \setminus f^{-1}(B)$
and
$\displaystyle f^{-1}\left(\bigcup_{i=1}^{\infty} B_i\right) = \bigcup_{i=1}^{\infty}f^{-1}(B_i)$,
$\displaystyle f^{-1}\left(\bigcap_{i=1}^{\infty}B_{i}\right) = \bigcap_{i=1}^{\infty}f^{-1}(B_{i}).$
It follows that $\mathcal M = \{ B \subset Y : f^{-1}(B) \in \mathcal{A}\}$ is a $\sigma$-algebra on $Y$. By assumption, $\mathcal{M} \supset \mathcal{G}$ and therefore $\mathcal{M} \supset \sigma(\mathcal{G}) = \mathcal{B}$, which implies that $f$ is measurable.
Now I got a few questions:
1) How do we know $\mathcal M$ is a $\sigma$-algebra on $Y$?
2) "By assumption, $\mathcal{M} \supset \mathcal{G}$" - why? $\mathcal M$ is just a sigma algebra on $Y$, one of many, it doesn't have to contain every subset of $Y$, so we don't know if it contains $\mathcal{G}$.
3) Why $\mathcal{M} \supset \sigma(\mathcal{G}) = \mathcal{B}$ implies that $f$ is measurable?
Please explain as simple as possible.

REPLY [2 votes]: 1) You need to check this. It's very straightforward, just use the definitions. 
2) $M$ is the sigma algebra of all sets such that their pre images are in $\mathcal{A}$ (i.e., $M = \{B \in Y: f^{-1}(B) \in \mathcal{A}\}$). You have that $f^{-1}(G) \in \mathcal{A}$ for all $G \in \mathcal{G}$, so clearly $\mathcal{G} \subset M$.
3) The definition of measurable states $f$ is measurable iff the pre-images of every set in $\mathcal{B}$ is in $\mathcal{A}$.$\sigma(\mathcal{G})=\mathcal{B} \subset M$. Recall that $M$ contains all sets such that their pre images are in $\mathcal{A}$ (by definition). Thus, we must have that the preimage of every set in $\mathcal{B}$ is in $\mathcal{A}$ and $f$ is measurable.