The completion of the Borel $\sigma$-algebra the same as the completion of the Lebesgue outer measure?

My study group and I were discussing this question today.

We can construct the Lebesgue measure using Caratheodory's extension theorem in the usual way:

• Given the function $F(x) = x$, we can construct a pre-measure $\mu_F$ associated with $F(x)$ defined on an algebra of "intervals" (Folland uses right-closed h-intervals);
• From this premeasure, we can induce an outer measure $\mu^*$ on the power-set of the reals;
• Using Caratheodory's extension theorem, the collection of $\mu^*$-measurable sets is a complete $\sigma$-algebra, and $\mu^*$ restricted to this $\sigma$-algebra is a complete measure.

This complete $\sigma$-algebra is in some sense a "big" structure; it is certainly larger than the Borel sets, and it must contain all of the Lebesgue measurable sets.

However, Folland provides a related Theorem:

Theorem 1.14 Let $\mathcal{A} \subset \mathcal{P}(X)$ be an algebra, $\mu_0$ a premeasure on $\mathcal{A}$, and $\mathcal{M}$ the $\sigma$-algebra generated by $\mathcal{A}$. There exists a measure $\mu$ on $\mathcal{M}$ whose restriction to $\mathcal{A}$ is $\mu_0$ -- namely, $\mu = \mu^*|_\mathcal{M}$, where $\mu^*$ is given by $$\mu^*(E) = \inf \left\{\sum_1^\infty \mu_0(A_j) : A_j \in \mathcal{A}, E \subset \bigcup_1^\infty A_j\right\}.$$

He then goes on to claim uniqueness. The proof of the theorem invokes Caratheodory, but there is a question that remains. I will try to make my thoughts clear:

• Caratheodory gives us a complete measure space, this structure is large.
• Theorem 1.14 as written tells us that we can use a premeasure on an algebra $\mathcal{A}$ to generate a measure on the $\sigma$-algebra generated by $\mathcal{A}$ -- this $\sigma$-algebra is not necessarily complete. In fact, this $\sigma$-algebra is just the Borel $\sigma$-algebra.
• We can complete the Borel $\sigma$-algebra to obtain the Lebesgue measurable sets.

However, is the completion of the Borel $\sigma$-algebra the same thing as we get by just applying Caratheodory's extension theorem to the Lebesgue outer measure to directly obtain a complete $\sigma$-algebra? Or is this a "bigger" structure than the completion of the Borel $\sigma$-algebra?

Let $\mathcal A$ be an algebra on a set $X$, $\mu_0$ a premeasure on $\mathcal A$, and $\mathcal M$ the $\sigma$-algebra generated by $\mathcal A$. Let $\mu^*$ be the Caratheodory outer measure induced by $\mu_0$, $\mu$ the restriction of $\mu^*$ on $\mathcal M$. Let $\mathcal C$ be the set of all $\mu^*$-measurable sets, $\bar{\mu}$ the restriction of $\mu^*$ on $\mathcal C$. Suppose $\mu_0$ is $\sigma$-finite, i.e. there exists a sequence $E_n, n = 1, 2, \cdots$ of members of $\mathcal A$ such that $X = \cup_{n=1}^{\infty} E_n$, $\mu_0(E_n) \lt \infty$ for all $n$. I claim that $(X, \mathcal C, \bar{\mu})$ is the completion of $(X, \mathcal M, \mu)$.
Let $\bar{\mathcal M}$ be the completion of $\mathcal M$ with respect to $\mu$. Since $\mathcal M\subset \mathcal C$, $\bar{\mathcal M}\subset \mathcal C$. Hence it suffices to prove that $\mathcal C \subset \bar{\mathcal M}$. Since $\mu_0$ is $\sigma$-finite, it suffices to prove that if $E\in \mathcal C$ and $\bar{\mu}(E) \lt \infty$, then $E\in \bar{\mathcal M}$. By the definition of $\bar{\mu}$, for each integer $n \ge 1$, there is a sequence $A_j, j= 1, 2, \cdots$ such that $\sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$, where $A_j \in \mathcal{A}$, $E \subset \bigcup_{j=1}^\infty A_j.$ Let $F_n = \bigcup_{j=1}^{\infty} A_j$. Then $E \subset F_n$ and $\mu(F_n) \le \sum_{j=1}^{\infty} \mu_0(A_j) \lt \bar{\mu}(E) + 1/n$. Let $F = \cap_{n=1}^{\infty} F_n$. Then $F\in \mathcal M$, $E \subset F$, and $\bar{\mu}(E) \le \mu(F) \le \mu(F_n) \lt \bar{\mu}(E) + 1/n$ for all $n\ge 1$. Hence $\bar{\mu}(E) = \mu(F)$. Similarly there exists $G\in \mathcal M$ such that $F - E \subset G$ and $\mu(G) = \bar{\mu}(F - E)$ = 0. Then $E = (F - G) \cup (E\cap G)$, $F - G \in \mathcal M$, and $E\cap G$ is a subset of the $\mu$-null set $G$. Hence $E \in \bar{\mathcal M}$. This completes the proof.
Now consider the family $\mathcal A$ of finite disjoint unions of intervals of the form $[a, b)$ or $(-\infty, c)$, where $-\infty \lt a\lt b\le \infty$ and $c\in \mathbb R$. It is elementary and well known that $\mathcal A$ is an algebra on $\mathbb R$ and there is a unique premeasure $\mu_0$ on $\mathcal A$ such that $\mu_0([a, b)) = b - a$ whenever $a$ and $b$ are finite. Then $\mathcal M$ and $\mathcal C$ defined above are the families of Borel sets and Lebesgue measurable sets in $\mathbb R$ respectively and you get the picture.
• Is there a counterexample if $\mu_0$ is not $\sigma$-finite? – wanderingmathematician Jun 27 '18 at 20:21
• Yes, there is. Take $X$ uncountable and the $\sigma$-algebra $\mathcal A$ of all subsets which are countable or have countable complement together with the counting measure. Then its completion is $\mathcal A$ since the only null set is $\emptyset$, but the $\sigma$-algebra $\mathcal C$ obtained by Caratheodory is the whole power set, $\mathcal P(X)$. – Staki42 Nov 20 '18 at 21:45