# Measurable mapping

A mapping $f$ of a measurable space $(X_1, \mathcal A_1)$ to a measurable space $(X_2, \mathcal A_2)$ such that $$f^{-1}(A) = \{x : f(x)\in A\} \in \mathcal A_1 \quad \text{ for each } A\in\mathcal A_2.$$ In the case where $\mathcal A_1$ is a $\sigma$-algebra and $(X_2, \mathcal A_2)$ is the real line with the $\sigma$-algebra $\mathcal A_2$ of Borel sets (cf. Borel set), the concept of a measurable mapping reduces to that of a measurable function (however, when $\mathcal A_1$ is only a $\sigma$-ring, the definition of a measurable function is usually modified in accordance with the requirements of integration theory). The superposition of measurable mappings is measurable. If $\mathcal A_1$ and $\mathcal A_2$ are rings and $f^{-1}(B)\in\mathcal A_1$ for each $B$ in some class of sets $B\in\mathcal A_2$ such that the ring generated by it is the whole of $\mathcal A_2$, then $f$ is measurable. The analogous assertions hold in the case of $\sigma$-rings, algebras and $\sigma$-algebras. If $(X_1, \mathcal A_1)$ and $(X_2, \mathcal A_2)$ are topological spaces with the $\sigma$-algebras of Borel sets, then every continuous mapping from $X_1$ to $X_2$ is measurable. Let $X$ be a topological space, let $\mathcal A$ be the $\sigma$-algebra of Borel sets and let $\mu$ be a finite non-negative regular measure on $\mathcal A$ (regularity means that $\mu(A) = \sup \{\mu(F) : F\subset A, F \text{ closed}\}$). Suppose further that $Y$ is a separable metric space, $\mathcal B$ is the $\sigma$-algebra of Borel sets, and let $f$ be a measurable mapping from $(X, \mathcal A)$ to $(Y, \mathcal B)$. Then for any $\varepsilon>0$ there is a closed subset $F\subset X$ such that $\mu(X\setminus F)<\varepsilon$ and $f$ is continuous on $F$ (Luzin's theorem).