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Let $L_d$ be the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}^d$.

By using Vitali's set $E \subseteq [0,1]$, I am looking for an example of $A \in L_2$ which is not in the product $\sigma$-algebra $L_1 \times L_1$.

I am also having trouble proving that $L_1 \times L_1 \subseteq L_2$. I can see that we can use $\mathcal{B}(\mathbb{R^2})=\mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R})$ and that the Lebesgue measure $\lambda_2$ on $(\mathbb{R^2},\mathcal{B}(\mathbb{R^2}))$ is identical to the product measure $\lambda_1 \times \lambda_1.$ Although I'm stuck afterwards.

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1 Answer 1

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  1. $L_1\otimes L_1$ is generated by $\mathcal{C}=\{A\times B:A,B\in L_1\}$. Since $\mathcal{C}\subset L_2$, $L_1\otimes L_1\subseteq L_2$.

  2. For a set $N\in L_1$ s.t. $N\ne \emptyset$ and $\lambda_1(N)=0$, the set $E\times N\in L_2$ ($\because E\times N\subset [0,1]\times N$ and $\lambda_2([0,1]\times N )=0$) but not in $L_1\otimes L_1$ ($\because$ for any $L_1\otimes L_1$-measurable set $A$, the sections $A^y=\{x\in \mathbb{R}:(x,y)\in A\}$ belong to $L_1$).

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