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While reading Bogachev's Measure Theory Vol I, I've stumbled upon a bit I don't follow: I don't understand the reasoning behind the very last sentence in the proof of Corollary 1.2.9 (see image below); namely, I don't see what follows from the second assertion nor do I understand the implication.

Would greatly appreciate some clarification!

Here's a link to a screencap of the relevant bitsCorollary 1.2.9

Thanks in advance.

EDIT: I think I've found the answer to this same question on another thread, so unless someone can think of a simpler solution i'd like to close this thread.

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Sai
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1 Answers1

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Consider the $\sigma$ algebra $\sigma(f^{-1}(\mathcal{F}))$. By the second assertion

$$\mathcal G:=\{ A \subset X | f^{-1}(A) \in \sigma(f^{-1}(\mathcal{F})) \}$$ is a $\sigma$-algebra.

As $\mathcal{F} \subset \mathcal G$ and $\mathcal G$ is a $\sigma$-algebra, you have

$$\sigma(\mathcal{F}) \subset \mathcal G \,.$$

Thus by the definition of $\mathcal G$ we have

$$f^{-1}(\sigma(\mathcal{F})) \subset \sigma(f^{-1}(\mathcal{F})) $$

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