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Let $\mathcal{L}(\mathbb{R})$ be the lebesgue-measurable set of $\mathbb{R}$, and $\mathcal{L}(\mathbb{R^2})$ the lebesgue-measurable set of $\mathbb{R^2}$.

First I shall show, that $\mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R})$ is a subset of $\mathcal{L}(\mathbb{R^2})$.

In a second part I shall show, that in fact they are not equal.

There is a hint, that the Lebesgue-measurable sets are the completion of the Borel sets.

So $\mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R})$ is the smallest $\sigma$-Algebra containing the set $\left\{ A \times B: A \in \mathcal{L}(\mathbb{R}), B \in \mathcal{L}(\mathbb{R}) \right\}$

But how to show all sets in this $\sigma$-Algebra are in fact lebesgue measurable in $\mathbb{R}^2$? I have no idea where to start. Any hint would be welcome.

1 Answers 1

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One characterization of $\mathcal{L}(\mathbb{R}^k)$ is the following: $ \mathcal{L}(\mathbb{R}^k)=\{C\cup N\mid C \in \mathcal{B}(\mathbb{R}^k),\; N\in\mathcal{N}^k\}, $ where $\mathcal{B}(\mathbb{R}^k)$ is the Borel-sets of $\mathbb{R}^k$ and $\mathcal{N}^k$ are the Lebesgue-nullsets in $\mathbb{R}^k$. As you suggest, it is enough to show that $A\times B\in \mathcal{L}(\mathbb{R}^2)$ for all $A,B\in \mathcal{L}(\mathbb{R})$. So let such $A$ and $B$ be given. Then they are of the form $ A=C_1\cup N_1,\quad B=C_2\cup N_2, $ where $C_2,C_2\in\mathcal{B}(\mathbb{R})$ and $N_1,N_2\in\mathcal{N}$. Now $ A\times B=(C_1\cup N_1)\times (C_2\cup N_2)=(C_1\times C_2)\cup (C_1\times N_2) \cup (N_1\times C_2)\cup(N_1\times N_2), $ above. Use the characterization above to conclude that $A\times B\in\mathcal{L}(\mathbb{R}^2)$.

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    Exactly. You can just prove the characterization yourself, if you haven't already.2012-06-27