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The version of Caratheodory's Theorem in Folland proceeds as follows:

If $\mu^*$ is an outer measure on $X$, the collection $\mathcal{M}$ of $\mu^*$-measurable sets is a $\sigma$-algebra, and the restriction of $\mu^*$ to $\mathcal{M}$ is a complete measure.

Suppose I have a $A \subset X$ and $\mu^*(A)=0$. Is this enough for $A \in \mathcal{M}$? I'm working on a problem and a solution uses this fact by saying "$\mu^*$ is complete" -- but this is a property of the measure, not the sigma algebra. Caratheodory's theorem says nothing about the completion of $\mathcal{M}$

The relevant question is here (part b, proving the reverse direction): Folland, Real Analysis problem 1.18

yoshi
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1 Answers1

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The answer is yes! Suppose that $\mu^*(A)=0$ for $A\subset X$. Then $\mu^*(E\cap A)\leq \mu^*(A)$, thus $\mu^*(E\cap A) = 0$ for every $E\subset X$. From this $$ \mu^*(E)\geq \mu^*(E\setminus A) = \mu^*(E\setminus A) + 0 = \mu^*(E\setminus A) + \mu^*(E\cap A). $$ The inequality $$ \mu^*(E)\leq \mu^*(E\setminus A) + \mu^*(E\cap A) $$ is obvious, thus we have $$ \mu^*(E) = \mu^*(E\setminus A) + \mu^*(E\cap A) $$ for every $E\subset X$, which means that $A\in\mathcal{M}$.

Albert
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