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Let $\mathcal{M}[0,1]$ be the $\sigma$-algebra of Lebesgue-measurable subsets of $[0,1]$ and $\lambda$ be the Lebesgue measure.

Are there two non-Lebesgue-measurable subsets $E_1$ and $ E_2$ of $[0,1]$ ($\not \in \mathcal{M}[0,1])$ and a countably-additive extension $\mu$ of $\lambda$ to the $\sigma$-algebra $\sigma(\mathcal{M}[0,1])\cup \{E_1, E_2\})$ such that

(1) $(\mu\upharpoonright_{\mathcal{M}_1})_* (E_2) \neq (\mu\upharpoonright_{\mathcal{M}_1})^* (E_2)$

(2) $(\mu\upharpoonright_{\mathcal{M}_2})_* (E_1) \neq (\mu\upharpoonright_{\mathcal{M}_2})^* (E_1)$

where

$\mathcal{M}_1=\sigma(\mathcal{M}[0,1]\cup \{E_1\})$ and $\mathcal{M}_2=\sigma(\mathcal{M}[0,1]\cup \{E_2\})$?

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1 Answer 1

up vote 1 down vote accepted

Take Lebesgue measure $\lambda^2$ on the square $[0,1] \times [0,1]$. This is isomorphic to Lebesgue measure on $[0,1]$. Now let $E_1$ be of the form $A \times [0,1]$, where $A$ is not $\lambda$-measurable; and let $E_2$ be $[0,1] \times A$.

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What is the extension you propose? Could you maybe elaborate, I'd be interested. Thanks. –  Sam Apr 9 '12 at 19:01
    
$(S, \mathcal{A}_1, \mu_1))$ is an extension of $(S, \mathcal{A}_2, \mu_2)$ if $\mathcal{A}_1 \supseteq \mathcal{A}_2$ and, for any event $E\in \mathcal{A}_2$, $\mu_1(E)=\mu_2(E)$. –  Joe Zhou Apr 10 '12 at 2:22

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