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Let $X$ be a set, and let $T$ be a $\sigma$-algebra on $X$.

Let $f: T \to [0,1]$ be a measure and let $\mu∗ f$ be the Carathéodory outer measure induced by $f$. Is it always the case that $T = T\mu∗ f$?

I wonder if we could use the following fact: you may freely use the fact that on the real line, there exists a Lebesgue measurable subset which is not Borel.

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    $\begingroup$ What is $T\mu*f$? $\endgroup$ Oct 13, 2021 at 23:24
  • $\begingroup$ Does $T\mu * f = \{S \subseteq X \mid [\mu * f](S) \neq \infty\}$? $\endgroup$ Oct 13, 2021 at 23:44
  • $\begingroup$ it means sigma-algebra of all measurable sets by μ∗f. μ∗f(A) =inf { sum of f(E) : {E} belongs to Covers of A) $\endgroup$
    – Asper
    Oct 14, 2021 at 0:21
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    $\begingroup$ The Question would be improved by incorporating that clarification into the body of the Question. As you can see from @StubbornAtom's edit, it is possible to post mathematical notation. You provided little in the way of context for the Question, and posts that only contain a problem statement (without context) may be closed for that reason. $\endgroup$
    – hardmath
    Oct 16, 2021 at 15:26

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The answer is no. Let $T$ be a sigma-algebra which is not complete with respect to $f$ (there are plenty of examples, ranging from trivial to $B_{\mathbb{R}}$). The sigma-algebra of Caratheodory measurable sets is complete with respect to $f$.

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  • $\begingroup$ Thank you so much. $\endgroup$
    – Asper
    Oct 15, 2021 at 1:35

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