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Let $(\Omega,\mathcal{A},\mu)$ be a measure space and $\mathcal B := \{A \in \mathcal A \mid \mu(A) < \infty\}$. We define a map $d_\mu:\mathcal A \times \mathcal A \to [0, \infty]$ by $d_{\mu}(A, B) := \mu(A \triangle B)$. Then $d_\mu$ becomes an extended metric when $\mathcal A$ is considered modulo the equivalence relation $\sim$ defined by $$ A \sim B \iff \mu(A \triangle B) = 0 \quad \forall A,B \in \mathcal A. $$

If there is a countable collection $(B_n) \subset \mathcal B$ such that $\mathcal A$ is generated by $(B_n)$, then $\mathcal B$ is dense in $\mathcal A$ w.r.t. the topology induced by $d_\mu$.

Are there some necessary and/or sufficient conditions such that $\mathcal B$ is dense in $\mathcal A$?

Akira
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    Hint: What is $d_\mu (A,B)$ if A has infinite measure, but $B$ has finite measure? What does this mean regarding your question? – PhoemueX Aug 06 '22 at 20:09
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    You wrote: "If there is a countable collection $(B_n) \subset \mathcal B$ such that $\mathcal A$ is generated by $(B_n)$, then $\mathcal B$ is dense in $\mathcal A$ w.r.t. the topology induced by $d_\mu$". Maybe I am missing something, but I don't think this is true. See counter-example below. – Ramiro Aug 06 '22 at 21:50
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    Consider $(\Bbb R,\mathcal{A},\mu)$, where $\mathcal{A}$ is the Borel $\sigma$-algebra and $\mu$ is the Lebesgue measure. Let $\mathcal B :={A\in\mathcal A\mid\mu(A)<\infty}$ and $\mathcal C :={[a,b]\mid a,b\in\Bbb Q}$. Then $\mathcal C\subseteq\mathcal B$, $\mathcal C$ is countable and $\mathcal A$ is generated by $\mathcal C$. On the other hand, if $A\in\mathcal A$ and $\mu(A)=\infty$, then, for all $B\in\mathcal B$, we have that $\mu(B)<\infty$, and so $d_{\mu}(A, B)=\mu(A\triangle B)=\infty$. So, $\mathcal B$ is not dense in $\mathcal A$ w.r.t. the topology induced by $d_\mu$. – Ramiro Aug 06 '22 at 21:50
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    In fact, in the general case, $\mathcal B$ is dense in $\mathcal A$ w.r.t. the topology induced by $d_\mu$ if and only if $\mathcal B = \mathcal A$, that is, $\mu$ is a finite measure. – Ramiro Aug 06 '22 at 21:56
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    Thank you so much @PhoemueX and @Ramiro! I got it. – Akira Aug 07 '22 at 00:31

1 Answers1

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As hinted by @PhoemueX and answered by @Ramiro, a set with infinite measure can not be approximated by a sequence of sets with finite measure. This is because $d_\mu (A, B) = \infty$ whenever $\mu(A) = \infty$ and $\mu(B) < \infty$. As such, my claim that

If there is a countable collection $(B_n) \subset \mathcal B$ such that $\mathcal A$ is generated by $(B_n)$, then $\mathcal B$ is dense in $\mathcal A$ w.r.t. the topology induced by $d_\mu$.

is plainly wrong.

Akira
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    Note that the topology (and also the metric) of $\mathcal A$ is that of a disjoint union of copies of $\mathcal B$, indexed by $\mathcal A/\mathcal B$. This will be separable if and only if $\mathcal A/\mathcal B$ is countable and $\mathcal B$ is separable, with the induced topology (which, I think, is equivalent to $L^1(\mu)$ being separable). – tomasz Aug 08 '22 at 10:16