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:

Source of confusion.

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!


I don't know why I didn't think of the obvious $X\cap A$...


1 Answer 1


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 $X$.


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