A have doubt.
Afirmation. All continuous function is Baire function.
Let $f:X\to \mathbb{R}$ continuous function. X compact hausdorff with $\mathcal{B}\ \sigma$-algebra of Baire. Let $\bigcap_{n=1}^{\infty} O_n\in\mathcal{B}(\mathbb{R})$ compact set. (sigma-algebra of Baire in $\mathbb{R}$) Now, $f^{-1}(\bigcap_{n=1}^{\infty} O_n)=\bigcap_{n=1}^{\infty}f^{-1}(O_n)$ compact set, and $f^{-1}(O_n)$ open sets. Therefore, $f^{-1}(\bigcap_{n=1}^{\infty} O_n)\in \mathcal{B}(X)$. Therefore $f$ is Baire function.
It is correct?
The Baire-$\sigma$-algebra is generated by closed $G_\delta$-sets. If $\bigcap_{i=1}^\infty O_i$ is closed with open $O_n$, then also $f^{-1}(\bigcap_{i=1}^\infty O_i)$ is closed. Moreover $f^{-1}(\bigcap_{i=1}^\infty O_i) = \bigcap_{i=1}^\infty f^{-1}(O_i)$. Again using the continuity we see that $f^{-1}(O_i)$ is open. Thus $f^{-1}(\bigcap_{i=1}^\infty O_i)$ is Baire-measurable.
Since we only need to prove measurability on a generating set-system, we get that $f$ is measurable according to the Baire-$\sigma$-algebras.