# Measure Theory Specially Carathéodory measure. and the sigma algebra induced by that [closed]

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.

• What is $T\mu*f$? Oct 13, 2021 at 23:24
• Does $T\mu * f = \{S \subseteq X \mid [\mu * f](S) \neq \infty\}$? Oct 13, 2021 at 23:44
• 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) Oct 14, 2021 at 0:21
• 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. Oct 16, 2021 at 15:26

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 $$f$$.