$K=\{E|f^{-1}(E)\in \mathcal{S} \}$ is a $\sigma$ algebra

Question

This is a problem I have seen while self studying measure theory:

Let $(X,\mathcal{S})$ be a measure space, and $E\in \mathcal{S} \otimes \mathcal{S}$, with $f(x) = (x,x)$ then $K=\{E|f^{-1}(E)\in \mathcal{S} \}$ is a $\sigma$ algebra.

How do I prove $K$ to a $\sigma$ algebra? I am unsure how it contains $\varnothing$, is closed under complement and countable unions.

score 2 · Accepted Answer · answered Nov 04 '20 at 01:42

2

Hint: Note that $\emptyset \in \mathcal{S}\otimes \mathcal{S}$ and $\emptyset \in \mathcal{S}$. Also, we have the following union and complement rules for pre-images: $$f^{-1}(E^{c}) = (f^{-1}(E))^{c} \quad \mbox{and} \quad f^{-1}\bigg{(}\bigcup_{n\in \mathbb{N}}E_{n}\bigg{)} = \bigcup_{n\in \mathbb{N}}f^{-1}(E_{n})$$

answered Nov 04 '20 at 01:42

JustWannaKnow

4,075

This makes a lot of sense! To clarify, $K$ is a sigma algebra on the set $X \times X$? – Jenny Liu Nov 04 '20 at 01:49
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No, it is a $\sigma$-algebra of subsets of $X\times X$ since its elements are also elements of $\mathcal{S}\otimes\mathcal{S}$. – JustWannaKnow Nov 04 '20 at 01:51
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That is what I meant to type, my apologies! – Jenny Liu Nov 04 '20 at 01:53
No problem! Then you are right! – JustWannaKnow Nov 04 '20 at 01:54
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@IamWill Since $K$ is a $\sigma$ algebra, why then is ${x\in X |(x,x)\in E } \in \mathcal{S}$? – XiaoMem24 Nov 04 '20 at 02:26

$K=\{E|f^{-1}(E)\in \mathcal{S} \}$ is a $\sigma$ algebra

1 Answers1