Let $f: X \to Y$, $\mathcal{C}$ be an arbitrary collection of subsets of $Y$, then $f^{-1}(\sigma(\mathcal{C})) \subset \sigma(f^{-1}(\mathcal{C}))$

Question

I have already read the solution to my question at https://math.stackexchange.com/a/1530785/81560, but unfortunately, the notation is confusing.

I am trying to prove:

Let $f: X \to Y$, and $\mathcal{C}$ be an arbitrary collection of subsets of $Y$. Then $f^{-1}(\sigma(\mathcal{C})) \subset \sigma(f^{-1}(\mathcal{C}))$.

I have already shown that $$\mathcal{A} = \{A \subset Y: f^{-1}(A) \in \sigma(f^{-1}(\mathcal{C}))\}$$ is a $\sigma$-algebra of subsets of $Y$. Now Yeh's Real Analysis, 3rd edition (p. 8) goes on to say

Clearly $\mathcal{A} \supset \mathcal{C}$

and this is the one part of the proof I don't understand. Why is this clearly true?

score 3 · Accepted Answer · answered Jun 05 '20 at 03:52

3

If $C\in\mathcal{C}$, then $f^{-1}[C]\in f^{-1}[\mathcal{C}]\subseteq\sigma(f^{-1}[\mathcal{C}])$, so $C\in\mathcal{A}$.

answered Jun 05 '20 at 03:52

Brian M. Scott

616,228

Let $f: X \to Y$, $\mathcal{C}$ be an arbitrary collection of subsets of $Y$, then $f^{-1}(\sigma(\mathcal{C})) \subset \sigma(f^{-1}(\mathcal{C}))$

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