I have the following Problem:
We take $(\Omega_1,A_1,\mu_1), (\Omega_1,A_1,\mu_1)$ be mesure spaces and $f: \Omega_1 \rightarrow \Omega_2$ be a mesurable map. The question is if the map $f$ is always $A_1^*, A_2^*$ mesurable where $A_i^*$ are the completions of the sigma algebras?
I have worked out the following solution but I'm not really sure if this works.
I take $\Omega_1=\Omega_2=\mathbb{N}$, $A_1=A_2=\{\emptyset, \mathbb{N}\}$ and $f(x)=x$ be the identity. Furthermore we take $\mu_1(X)=|X|$ if $X$ is finite and otherwise $\infty$ and $\mu_2(X)=0$. We remark that these are indeed measures and f is mesurable on these sigma algebras. But now I get thet $A_2^*=P(\mathbb{N})$ and $A_1^*=\{\emptyset,\mathbb{N}\}$. Since $\{1\}\in A_2^*$ we have $$f^{-1}(\{1\})=\{1\} \notin A_1^*$$ which shows that f is not mesurable on the completitions.
But does this work?
Sorry there are some guys who voted my question, does this mean to you also that it works or is it only because you liked the question?