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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

user123234
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2 Answers2

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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.

Adayah
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  • Nice easy answer! – Wreck it Ralph Oct 17 '21 at 17:25
  • Sorry if I ask here again, but is there a general rule, when such a map is always measurable on their completitions? – user123234 Oct 17 '21 at 18:20
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    One such sufficient condition is: the preimage of every $\mu_2$-null set is a $\mu_1$-null set. This easily follows from the characterization of the completion: $$\mathcal{A}^* = { A \triangle N : A \in \mathcal{A}, N \subseteq B \text{ for some } B \in \mathcal{A} \text{ with } \mu(B) = 0 }$$ – Adayah Oct 17 '21 at 18:42
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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.

Wreck it Ralph
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    but is the completition of the borel sigma algebra equal to the lebesgue sigma algebra? And isn't there a simpler example because we have never heard about the vitali set – user123234 Oct 17 '21 at 15:24
  • Yes it is equal. The Vitali set is a very nice construction of a nonmeasurable set and it is the reason of why people work with sigma algebras and not power sets of subsets: give a look to it and try to understand it, it is a must-have topic in measure theory. – Wreck it Ralph Oct 17 '21 at 17:00