How do I find an example s.t. a function is not mesurable on the completions of the simga-algebras.

Question

I have the following problem:

Let $(\Omega_1,A_1,\mu_1),(\Omega_2,A_2,\mu_2)$ be measure spaces and $f:\Omega_1\rightarrow \Omega_2$ be a $(A_1,A_2)$-mesurable map. We denote by $A_1^*,A_2^*$ the completions of the $\sigma$-algebras $A_1,A_2$.Is the map f always $(A_1^*,A_2^*)$-measurable. Give an example.

My claim was that it's not always measurable, but I can't find an example. My Idea was that the sigma algebra on the domain is smaller then the one of the codomain. Maybe in this case we get that if we complete both, one do not contail the preimages of nullsets of the other. But I don't know if this works.

Thanks for your help

Answer 1

The map need not be $(A_1^*, A_2^*)$-measurable.

Take $\Omega_1 = \Omega_2 = \mathbb{N}$, $A_1 = A_2 = \{ \varnothing, \mathbb{N} \}$ and $f = \mathrm{id}$. Also let $\mu_2 = 0$ and define $\mu_1$ so that $\mu_1(\mathbb{N}) = 1$.

Then obviously $f$ is $(A_1, A_2)$-measurable. But since $\mu_2$ is zero, $A_2^* = \mathcal{P}(\mathbb{N})$ while $A_1^* = A_1$ so clearly $f$ is not $(A_1^*, A_2^*)$-measurable.

Answer 2

Denote $\mathcal B$ and $\mathcal L$ denote the Borel and Lebesgue $\sigma$-algebra, respectively.

Consider $g :\mathbb R\to\mathbb R^2$ given by $g(x) = (x,0)$ which is $(\mathcal B$-$\mathcal B^2)$-measurable since it is continuous, so the preimage of a Borel subset of $\mathbb{R}^2$ will be a Borel subset of $\mathbb{R}$. Thus $g$ is Borel to Borel measurable.

But it is not $(\mathcal L$-$\mathcal L^2)$-measurable: let $N\subseteq[0,1]$ be a Lebesgue nonmeasurable set, which always exists (see Vitali set). Then $M\times\{0\}$ is a Lebesgue measurable subset of $\mathbb{R}^2$ because the Lebesgue measure is complete, $M\times\{0\}\subseteq[0,1]\times\{0\}$, and $[0,1]\times\{0\}$ has Lebesgue measure zero. Then $g^{-1}(M\times\{0\})=M$ shows that $g$ is not Lebesgue to Lebesgue measurable.