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Broadly speaking, I am not sure how to show that a particular set is part of a product $\sigma$-algebra.

In particular, I am trying to show that $S = \{(x,y) \in \mathbb{R} \times \mathbb{R}: x = y\}$ is in $\mathcal{P}(\mathbb{R}) \times \mathfrak{M}$ (as part of a larger question about product measure spaces).

Here, $\mathfrak{M}$ denotes the Lebesgue measurable subsets of $\mathbb{R}$ and $\mathcal{P}(\mathbb{R})$ is the power set of the reals.

I’m not sure how to proceed. Any tips on how to solve the problem generally (or related exercises) are appreciated!

TuringTester69
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  • You cannot show that $S$ is in $\mathscr{P}(\mathbb{R}) \times \mathfrak{M}$ as it is clearly not a product set, but you can show it is in $\mathscr{P}(\mathbb{R}) \otimes \mathfrak{M}$, which is the $\sigma$-algebra generated by that set. – Henno Brandsma Dec 02 '18 at 08:59

2 Answers2

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$\mathscr{B}(\mathbb{R}^2)=\mathscr{B}(\mathbb{R})\otimes \mathscr{B}(\mathbb{R})\subseteq\mathcal{P(\mathbb R)}\otimes \mathscr{B}(\mathbb{R})\subseteq \mathcal{P(\mathbb R)}\otimes \frak M $ and the diagonal is contained in $\mathscr{B}(\mathbb{R}^2)$.

Matematleta
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Is a stronger fact that $S$ is in the sigma-algebra generated by the topology (the borelians)-see @Matematleta answer.

For proving this you just prove that $S$ is closed. You can do this in many ways... for example, consider the continuous real function $h(x)=x-y$, then $S=h^{-1}(\{0\})$ and hence $S$ is closed.

Therefore, $S^c$ is a open subset of $\Bbb{R}^2$. And there is a very known result that says that every open set of a separable topological space is at most countably union of open balls (the sets which determines the topology), here you can see many ways for proving this fact on $\mathbb{R}$. So you conclude that $S^c=\bigcup\limits_{n\in\Bbb{N}}B_n$ with $B_n$ elements in the topology $\tau$, so $S^c \in \sigma(\tau)=\mathcal{B}(\Bbb{R}^2)$ and then you can conclude what you want!

Marcelo
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