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The question comes from a statement in the last paragraph of the proof of Theorem 1.4 on page 11 of Lieb and Loss's Analysis, GSM Volume 14.

Let $\Omega$ be a set, $\mathcal{A}$ an algebra of subsets of $\Omega$ and $\Sigma$ the smallest sigma-algebra that contains $\mathcal{A}$.

Now, according to the book, for any $A_0\in\mathcal{A}$, $A_0\cap\Sigma$ is a sigma-algebra on $A_0$ which is the smallest one that contains the algebra $A_0\cap\mathcal{A}$. The notation should be clear; \begin{equation*} A_0\cap\mathcal{A} = \left\{A_0\cap E: E\in\mathcal{A} \right\} \end{equation*} and \begin{equation*} A_0\cap\Sigma = \left\{A_0\cap E: E\in\Sigma \right\}. \end{equation*}

I don't find it difficult to show that

  1. $A_0\cap\Sigma$ is a sigma-algebra on $A_0$.
  2. $A_0\cap\mathcal{A}$ is an algebra.

However, it is not clear to me that $A_0\cap\Sigma$ is the the smallest sigma-algebra on $A_0$ that contains the algebra $A_0\cap\mathcal{A}$. I have thought along the lines of Theorem 1.3 (Monotone class theorem) but didn't make any progress. Could anyone please offer some insight?

khalatnikov
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  • What you want to prove is $\sigma(A_0 \cap \mathcal{A}) = A_0 \cap \Sigma$ and you have $\sigma(A_0 \cap \mathcal{A}) \subset A_0 \cap \Sigma.$ What about trying the set of $E$ in $\Sigma$ such that $A_0 \cap E$ is in the sigma algebra $\sigma(A_0 \cap \mathcal{A})$? – William M. Dec 28 '16 at 02:44
  • @WillM. Thank you for the suggestion. Any set in $A_0\cap\Sigma$ would be of the form $A_0\cap E$ for some $E\in\Sigma=\sigma(\mathcal{A})$. Hence $A_0\cap E\in A_0\cap\Sigma=A_0\cap\sigma(\mathcal{A})=\sigma(A_0\cap\mathcal{A})$, from which we can conclude that $A_0\cap\Sigma\subset\sigma(A_0\cap A)$. I think it still remains to show that $A_0\cap\sigma(\mathcal{A})=\sigma(A_0\cap\mathcal{A})$. – khalatnikov Dec 28 '16 at 09:58

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Let $\mathcal B$ be a $\sigma$-algebra (on $A_0$) containing $A_0\cap\mathcal A$. Define $\mathcal D:=\{D\subset\Omega:A_0\cap D\in\mathcal B\}$. Clearly $\mathcal D\supset\mathcal A$, and you can check that $\mathcal D$ is a $\sigma$-algebra (on $\Omega$). It follows that $\mathcal D\supset \sigma(\mathcal A)=\Sigma$. That is, $A_0\cap\Sigma\subset\mathcal B$, yielding the desired minimality of $A_0\cap\Sigma$.

John Dawkins
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