Unusual definition of function measurability

Question

I came across this definition "A function $f:\Omega_1 \rightarrow \Omega_2$ is measurable if it is $(\Sigma(\Omega_1), \Sigma(\Omega_2))$-measurable", where $\Omega_{1, 2}$ stands for Borel space and $\Sigma(\Omega_{1, 2})$ is a Borel $\sigma$-algebra. I do not understand the "$(\Sigma(\Omega_1), \Sigma(\Omega_2))$-measurable" part. What does it mean?? There is no additional elaboration on this definition as if it is some commonly known fact. The only definition of the function measurability I know is the one presented as follows: Let $(X, \Sigma, \mu)$ be a measure space. Then function $f(x)$, $x\in X$, is measurable if for any $c\in R$ the set $\{x\in X|f(x)<c\}$ is measurable.

Answer 1

If $\Sigma_1$ is a $\sigma$-algebra on $\Omega_1$, and similarly $\Sigma_2$ is a $\sigma$-algebra on $\Omega_2$, the statement "$f\colon \Omega_1\to \Omega_2$ is measurable" means that for each $E\in\Sigma_2$, $f^{-1}(E)\in\Sigma_1$. This is analogous to the general definition of a continuous function $f\colon (X,\tau_1)\to (Y,\tau_2)$ of topological spaces, which says $f^{-1}(U)\in\tau_1$ for each $U\in\tau_2$.

In the context of functions taking values in $\mathbb R^n$, the $\sigma$-algebra we take on $\mathbb R^n$ is commonly the Borel $\sigma$-algebra, in which case measurability can equivalently be reformulated as asking $$ f^{-1}(\prod_{i=1}^n(-\infty,a_i])\in\Sigma_1\quad\text{ for each $(a_1,\dots,a_n)\in\mathbb R^n$}. $$