# Let $\mathfrak{C}$ be an arbitrary collection of subsets of a set $X$ and $A \subset X$. Does $\sigma({\mathfrak{C}})$ contain A?

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 $$X\cap A$$...

Nowhere does the book say that $$A \in \sigma (\mathbb C)$$. Is says that $$\sigma (\mathbb C) \cap A$$ is a sigma algebra of subsets of $$A$$. First note that this contains $$A$$ because $$A=X\cap A$$ and $$X \in \sigma (\mathbb C)$$. If $$A_n \in \sigma (\mathbb C) \cap A$$ for $$n=1,2,...$$ then $$A_n= C_n \cap A$$ with $$C_n \in \sigma (\mathbb C)$$ and $$\cap_n A_n =\cap_n C_n \cap A$$ so $$\cap_n A_n \in \sigma (\mathbb C) \cap A$$. can you verify closure under complements? Remenber to take complements in $$A$$, not in $$X$$.