# Largest $\sigma$-algebra on which an outer measure is countably additive

If $m$ is an outer measure on a set $X$, a subset $E$ of $X$ is called $m$-measurable iff $$m(A) = m(A \cap E) + m(A \cap E^c)$$ for all subsets $A$ of $X$.

The collection $M$ of all $m$-measurable subsets of $X$ forms a $\sigma$-algebra and $m$ is a complete measure when restricted to $M$.

Is $M$ the largest $\sigma$-algebra on $X$ on which $m$ is a measure (i.e., on which $m$ is countably additive)? If not, what is?

Is $M$ the largest $\sigma$-algebra on $X$ on which $m$ is a complete measure? If not, what is?

I am especially interested in the case when $X$ is $\mathbb{R}$ or $\mathbb{R}^n$ and $m$ is the Lebesgue outer measure. In this case $M$ is the Lebesgue $\sigma$-algebra.

ADDED:

Julián Aguirre (thanks!) has shown in his response below that the answer to the first question is yes when $X$ is $\mathbb{R}^n$ and $m$ is the Lebesgue outer measure. Hence the answer to the second question in this situation is also yes.

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## 1 Answer

The answer to the first question (and the one in the title) is yes when $M$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}^n$. Suppose $N$ is another $\sigma$-algebra such that $M\subsetneq N$. Then there exists $E\in N$ non measurable. Since $E=\cup_{k=1}^\infty (E\cap\{x\in\mathbb{R}^n:|x|\le k\})$, for at least one $k$ the set $E\cap\{x\in\mathbb{R}^n:|x|\le k\}$ is not measurable. Thus, we may assume without loss of generality that $E$ is bounded, and in particular, $m(E)<+\infty$.

Since $E$ is not measurable, there exists $\epsilon>0$ such that if $G$ is an open set and $E\subset G$, then $m(G\setminus E)\ge \epsilon$. (This follows from an equivalent definition of measurable set; cf. Proposition 15 on p.63 of Royden's Real Analysis, 3 ed.)

Now, since $m(E)=\inf\{m(G):G\text{ open, }E\subset G\}$, there exists an open set $O$ such that $E\subset O$ and $m(E)\ge m(O)-\epsilon/2$. Then $O=(O\setminus E)\cup E$ but $$m(O\setminus E)+m(E)\ge\epsilon+ m(O)-\epsilon/2>m(O),$$ and hence $m$ is not additive on $N$.

For the general case it is easy to see that if $N$ is another $\sigma$-algebra such that $M\subsetneq N$, then there exists $A\subset X$ such that $m$ is not additive on the $\sigma$-algebra generated by $N\cup\{A\}$, but I have not been able to show that it is not additive on $N$.

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