I'm struggling with this exercise. I have no idea on how to proceed. Do you have any suggestion?
Let $(Ω_1,A_1,μ_1)$ and $(Ω_2,A_2,μ_2)$ be measure spaces and $f : Ω_1 \to Ω_2$ a $(A_1,A_2)$-measurable map. We denote by $\overline{A}_1$ and $\overline{A}_2$ the completions of the σ-algebras $A_1$ and $A_2.$
(a) Is the map $f$ always $(\overline{A}_1, \overline{A}_2)$-measurable? Example?
(b) Formulate and prove a condition which implies that f is $(\overline{A}_1, \overline{A}_2)$- measurable.
Thank you!!
Partial answer. In a) the answer is negative.
Denote by $K(x)$ the Cantor function, put $\Omega_1 = \Omega_2 = [0,1]$, $A_1$ = sigma-algebra of all Lebesgue-measurable sets, $A_2$ = Borel sigma-algebra, $\mu_1 = \mu_2 = \mu=$ Lebesgue measure. Denote closed Cantor set by $E$. Let $g(x) = \frac{K(x)+x}{2}$ - continious and strictly monotone function (bijection). Then $[0,1] \backslash E = $ open Cantor set = $\cup_{n \ge 1} (a_n, b_n)$ and $\sum_{n \ge 1}(b_n-a_n) = 1$. Further $g([0,1] \backslash E) = \sqcup_{n \ge 1} (g(a_n), g(b_n))$. Thus $\mu \bigl( g([0,1] \backslash E) \bigr) = \sum_{n \ge 1} \frac{b_n-a_n}{2} = \frac12.$ But $g([0,1]) = [0,1]$ and hence $\mu(g(E)) = 1 - \frac12$. There exists $B \subset g(E)$ such that $B$ is not Lebesgue measurable set. We have $g^{-1}(B) \subset E$ and hence $g^{-1}(B)$ is Lebesgue measurable with measure $0$.
Denote $g^{-1}$ by $f$: it exists and is continious and strictly monotone. Put $C = f(B)$. We know that $C$ is Lebesgue measurable and that $f^{-1}(C) = f^{-1} \bigl( f(B) \bigr) = B$ is not Lebesgue measurable.
We got $f$: $(\Omega_1, A_1, \mu_1) \to (\Omega_2, A_2, \mu_2)$ such that $f$ is continious and hence measurable. Moreover, $\overline{A_1} = A_1$, $\overline{A_2} = A_1$ and $f$ is not measurable as a map from $(\Omega_1, \overline{A_1}, \mu_1)$ to $(\Omega_2, \overline{A_2}, \mu_2)$ because $C \in \overline{A_2} = A_1$ and $f^{-1}(C) \notin \overline{A_1} = A_1$.
Hence even continuity + strictly monotonicity of function $f$ is not sufficient condition in b).