We know that a measurable space is defined as a tuple $(X,\mathcal A)$, where $X$ is a set and $\mathcal A$ is a $\sigma$-algebra defined on $X$. A $\sigma$-algebra is traditionally defined as a collection of subsets such that the whole set is in collection and it is closed under complements and countable union. A function $f:(X,\mathcal A)\to (Y,\mathcal B)$ is called measurable if for every $U\in \mathcal B,\ f^{-1}(U)\in\mathcal A$. We note that this definition uses the pre-image as a central notion.
An analogous construction is a topology $\tau$ on a set $X$, where $\tau$ is defined as a collection of subsets of $X$ such that it contains the whole set and the empty set and is closed under arbitrary union and finite intersection. A function $f:(X,\tau)\to (Y,\tau')$ is called continuous if for every $U\in \tau',\ f^{-1}(U)\in\tau$. We again note that this definition uses the pre-image as a central notion.
But in the case of topology, there exists an equivalent characterization using the so called Kuratowski closure operator $\bf c$ in terms of which continuous functions are defined as a function $f:(X,\mathbf{c})\to(Y,\mathbf{c}')$ such that $f(\mathbf c(U)) \subseteq \mathbf c'(f(U))$ for all $U\subseteq X$. We note that this definition does not use pre-images and is equivalent to the traditional definition of continuous functions.
Is there a similar reformulation of $\sigma$-algebras such that measurable functions also does not use pre-images in it's definition? In fact, is there a reformulation such that measurable functions can be defined as exactly those functions such that $f(g(U))\subseteq g'(f(U))$ for all $U\subseteq X$, where $g,g'$ are functions from power set of $X,Y$ to power set of $X,Y$ respectively?
