Alannah Short

2022-06-26

Let $(X,\mathcal{A})$ y $(Y,\mathcal{B})$ measure spaces. Let $R:=R(\mathcal{A},\mathcal{B})$ the collection of measurable rectangles $R:=\{A\times B:A\in \mathcal{A},B\in \mathcal{B}\}.$. Prove that the algebra of subsets of $X\times Y$ generated by $R$ is the collection of finite unions of elements of $R$. I have a hint and it is in which I call $C$ to be the collection of finite unions of elements of $R$, prove that $C$ is in the algebra generated by $R$, then prove that $C$ is an algebra that contains $R$.

EreneDreaceaw

Beginner2022-06-27Added 20 answers

You only need to prove that the set $\mathcal{A}$ of finite disjoint unions of elements of $R$ is an algebra.

Note that $R$ has the following properties:

$\mathrm{\varnothing}\in R,$

$E,F\in R\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}E\cap F\in R,$

$E\in R\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{c}\text{is a finite disjoint union of members of}R.$

Any set $R$ that satisfies the above three hypotheses has the property that the algebra generated by $R$ is the collection of finite disjoint unions of elements of $R$. This is not difficult to verify.

Note that $R$ has the following properties:

$\mathrm{\varnothing}\in R,$

$E,F\in R\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}E\cap F\in R,$

$E\in R\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{E}^{c}\text{is a finite disjoint union of members of}R.$

Any set $R$ that satisfies the above three hypotheses has the property that the algebra generated by $R$ is the collection of finite disjoint unions of elements of $R$. This is not difficult to verify.

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