If $f: (X, M) \to (Y, \tau)$ is measurable, then for all Borel sets $B \subset Y$ it is true that $f^{-1}(B) \in M$

Question

I am lost with this proof, the problem is as follows:

Let $(X,M)$ be a measurable space, $(Y, \tau)$ be a topological space, and $f:X\to Y$ be measurable. Then, for any Borel set $B \subset Y \Rightarrow f^{-1}(B) \in M$.

I know that, since $f$ is measurable, then it must be true that for any set $U \in \tau \Rightarrow f^{-1}(U) \in M$ by the definition of measurable function. But this seems too simple and I am almost sure I am missing steps.

Any help is appreciated :)

Answer 1

Actually, this is just the definition of measurability but if your definition says inverse images of open sets are in $M$ then here is a proof: the collection of all Borel sets $B$ in $Y$ such that $f^{-1}(B) \in M$ is a sigma algebra containing all open sets, hence all Borel sets.

Answer 2

Let $\mathcal{B}_f$ be the family $\{B \subseteq Y: f^{-1}[B] \in M\}$.

Then $\mathcal{B}_f$ is a $\sigma$-algebra as $f^{-1}$ preserves unions, complements etc. and $M$ is a $\sigma$-algebra. [insert your own proof here.]

By your definition of measurable function: $\tau \subseteq \mathcal{B}_f$.

The Borel sigma algebra $\mathrm{Bor}_Y$ of $Y$ is by definition the smallest $\sigma$-algebra that contains $\tau$. $\mathcal{B}_f$ is one $\sigma$-algebra that contains $\tau$ so by the minimality $\mathrm{Bor}_Y \subseteq \mathcal{B}_f$ and this means by definition of the latter that $f^{-1}[B]\in M$ for all Borel subsets $B$ of $Y$.