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**Focus** As B is a Borel set, it is the intersection of some open sets (countable union of open sets is also open, all we are missing is the countable intersection), so $\displaystyle B=\bigcap G_n$.

You definitely can't say that. Borel subsets can be much more complicated than intersections of open sets.

The Borel $\displaystyle \sigma$-algebra is defined as the smallest $\displaystyle \sigma$-algebra containing the open subsets. That's the only property we can use to prove that a set is in the Borel $\displaystyle \sigma$-algebra. Hence the proof can't be direct. We know that $\displaystyle f^{-1}(O)\in\mathcal{B}$ for any open subset $\displaystyle O$ (because $\displaystyle f^{-1}(O)$ is open, too). And we want to generalize this to any borelian subset instead of just open subsets.

The proof goes this way: let

$\displaystyle \mathcal{C}=\{B\in\mathcal{B}|f^{-1}(B)\in\mathcal{B}\}$.

Then $\displaystyle \mathcal{C}$ is a subset of $\displaystyle \mathcal{B}$, it contains the open subsets, and it is easy (but you really should write it down) to prove that $\displaystyle \mathcal{C}$ is a $\displaystyle \sigma$-algebra. Therefore, by definition of the borelian $\displaystyle \sigma$-algebra, $\displaystyle \mathcal{B}\subset\mathcal{C}$, hence $\displaystyle \mathcal{C}=\mathcal{B}$. This is exactly what we needed: for any $\displaystyle B\in\mathcal{B}$, $\displaystyle f^{-1}(B)\in\mathcal{B}$.