Measurable function and Borel set We define measurable function from a measure space to a topological space as space which pulls back open sets to measurable sets. How can we prove measurable functions pulls back Borel set also to measurable sets. 
where Borel sets are elements in sigma algebra generated by topology
 A: Define
$$
\Sigma_f := \{M \in \mathcal{B} \mid f^{-1}(M) \in \mathcal{A} \},
$$
where $\mathcal{A}$ is the chosen $\sigma$-algebra on your measurable space.
Show that $\Sigma_f$ is a $\sigma$-algebra. Why does that help you?
EDIT: The same proof shows that it suffices to show $f^{-1}(M) \in \mathcal{A}$ for all $M \in \mathcal{M}$ for any family $\mathcal{M}$ of subsets to conclude that $f$ is measurable w.r.t $\mathcal{A}$ and the generated $\sigma$-algebra $\sigma(\mathcal{M})$.
A: Let $f:X\rightarrow Y$ be a function and let $\mathcal{O}\subseteq\wp\left(Y\right)$. In your case $\mathcal O$ is the topology.
It comes to proving that $\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)=f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$
where in general $\sigma\left(\mathcal{V}\right)$ denotes the $\sigma$-algebra
generated by collection $\mathcal{V}$. 


*

*If $\mathcal{A}$ is a $\sigma$-algebra on $X$ then $\left\{ A\in\wp\left(Y\right)\mid f^{-1}\left(A\right)\in\mathcal{A}\right\} $
is a $\sigma$-algebra on $Y$.


Filling in $\mathcal{A}=\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)$
we find this collection to be a $\sigma$-algebra containing $\mathcal{O}$.
Then it also contains $\sigma\left(\mathcal{O}\right)$. This is exactly
the statement that $f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)\subseteq\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)$.


*

*If $\mathcal{B}$ is a $\sigma$-algebra on $Y$ then $f^{-1}\left(\mathcal{B}\right)$
is a $\sigma$-algebra on $X$.


Filling in $\mathcal{B}=\sigma\left(\mathcal{O}\right)$ we find $f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$
to be a $\sigma$-algebra with $f^{-1}\left(\mathcal{O}\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$
and consequently $\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$.
