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Let $X\neq\emptyset$ and let $\mu:P(X)\to[0,\infty]$ be an outer measure. Recall that, a set $A\subseteq X$ is $\mu$-measurable if $$ \mu(B)=\mu(A\cap B)+\mu(B\setminus A), \text{ for all }B\subseteq X. $$ Let $M(X,\mu):=\{A\subseteq X:A\text { is }\mu-\text{measurable}\}$ be the set of all $\mu$-measurable sets. We know that $M(X,\mu)$ is a $\sigma$-algebra on $X$.

QUESTION:

My question is about the converse. Given any $\sigma$-algebra on $X$, does there exist an outer measure $\mu$ on $X$ such that $$ \boxed{\sigma=M(X,\mu)?} $$ If not, is there any necessary and sufficient condition on $\sigma$ that ensures the existence of an outer measure $\mu$ on $X$ such that $ \sigma=M(X,\mu)? $

YCor
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Tatin
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    The constructed measure will be complete. So the Borel sets on the real line are a counterexample. – Michael Greinecker Nov 14 '22 at 09:57
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    @MichaelGreinecker Thank you for your comments. How do we know that the set of all Borel sets on R is not complete with respect to some outer measure? What we know is that it is not complete with respect to the Lebsegue measure. – Tatin Nov 14 '22 at 10:34
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    There are results about such spaces that imply there will be an uncountable Borel set with measure zero, at least under a $\sigma$-finite measure. An uncountable Borel set has the cardinality of the continuum. Butany more subsets. – Michael Greinecker Nov 14 '22 at 10:41
  • And there are only continuum many Borel sets. – Michael Greinecker Nov 14 '22 at 10:55
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    Related question: https://mathoverflow.net/q/87838/1946 – Joel David Hamkins Nov 14 '22 at 12:00

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