My preliminary working suggests that this is false but the book I'm following (Real Analysis : Theory of Measure and Integration by James Yeh) suggests it's true. From experience, it's more likely that the author is correct and so I'm asking where did I go wrong?
For reference, here's the quoted text where the relevant part is highlighted:
I decided to verify myself (below) if indeed $\sigma(\mathfrak{C})\cap A$ is a $\sigma$-algebra of subsets of $A$:
$\sigma(\mathfrak{C}) \cap A = \{F \cap A: F\in \mathfrak{C}\}$
My first thought was to check if $A \in \mathfrak{C}$ because this implies $A \in \sigma(\mathfrak{C}) \cap A $ which fulfils one of the conditions of being a $\sigma$-algebra. I decided to investigate a naive case to see if this is always true below:
let $X = \{1,2,3\}$, $\mathfrak{C}$ = $\{\{1\}\} \subset 2^{X}$, $A = \{1,3\} \subset X$.
Then clearly, $A = \{1,3\} \notin \sigma(\mathfrak{C}) = \sigma(\{\{1\}\}) = \{\emptyset, \{1,2,3\}, \{1\}, \{2,3\}\}$.
What I got from that is that it's not guaranteed that a subset $A$ of $X$ would be contained in a $\sigma$-algebra of subsets of an arbitrary collection $\mathfrak{C}$ of $X$.
And this is where I've come stuck. Any help would be appreciated!
Edit:
I don't know why I didn't think of the obvious $X\cap A$...