Let $X$ be a measurable space.(i.e. $X$ is equipped with sigma algebra $R$.) Furthermore, note that $R$ really defines a sheaf of sigma algebra via $R\cap U$ for any open $U\subset X$ where intersection is obvious choice on elements.) $f:X\to Y$ is a measurable map. Then $f_\star R=\{E\subset Y\vert f^{-1}(E)\in R\}$ defines sigma algebra on $Y$ which also forms a sheaf.
$\textbf{Q:}$ Since this looks like pushing forward the sheaf, I would certainly hope there is inverse image functor $f^{-1}$.(i.e. $f:X\to Y$ as "morphism" s.t. $Y$ is equipped with some $\sigma$ algebra $R_Y$ with $f^{-1}(R_Y)$ defines sigma algebra over $X$.) However, in algebraic geometric setting, I need sheafification procedure going on. It seems that I do not need it here? I want to say $f^{-1}(R_Y)=\{f^{-1}(U)|U\in R_Y\}$ defines $\sigma$ algebra over $X$ Is this even correct?