Understanding Pre-Measures and Extensions of Measures in Folland

I've been reading through Folland's section on outer measures and premeasures, and I really want to understand the bigger picture as I feel as if though I'm losing sight of the bigger picture. My main problem is with Theorem 1.14 and its proof but first I introduce some important concepts with my questions on them.

1.11 Carathéodory's Theorem: If $$\mu^*$$ is an outer measure on X, the collection $$M$$ of $$\mu^*$$ measurable sets is a $$\sigma$$ algebra, and the restriction of $$\mu^*$$ to $$M$$ is a complete measure.

$$\textbf{My questions: Does this mean on M, \mu^* is countably additive}$$ $$\textbf{so that it is indeed a measure?}$$ $$\textbf{Are all subsets of null sets also contained in M, to be complete?}$$ $$\textbf{Folland to prove completeness shows that if \mu^{*}(A)=0,\mbox{then A \in M}, but}$$ $$\textbf{shouldn't he show that subsets of null sets are contained in M, not just that null sets are}$$ $$\textbf{are contained in M?}$$ This is what he does:

1.13: If $$\mu_0$$ is a premeasure on $$A \subset P(X)$$, where $$A$$ is an algebra and $$u^*$$ is defined by: $$u^{*}(E)=\mbox{inf}$${$$\sum \limits_{1}^{\infty} \mu_0(A_j): A_j\in A, E \subset \bigcup\limits_{i=1}^{\infty}A_j$$}, then:

a. $$\mu^{*}|_A=\mu_0$$

b. every set in $$A$$ is $$\mu^{*}$$ measurable.

Does 1.13a just simply mean mean that on A for E $$\subset A,u^*{E}=\mu_0(E)$$?

Theorem 1.14: Let $$A \subset P(X)$$ be an algebra, $$\mu_0$$ a premeasure on $$A$$, and $$M$$, the $$\sigma-algebra$$ generated by $$A$$. There exists a measure $$\mu$$ on $$M$$ whose restriction to $$A$$ is $$\mu_0$$-namely,$$\mu|_M=\mu^{*}$$, where $$\mu^{*}$$ is given by

$$u^{*}(E)=\mbox{inf}$${$$\sum \limits_{1}^{\infty} \mu_0(A_j): A_j\in A, E \subset \bigcup\limits_{i=1}^{\infty}A_j$$}. If $$\upsilon$$ is another measure on $$M$$ that extends $$u_0$$, then $$\upsilon \leq \mu(E)$$ for all $$E \in M$$, with equality when $$\mu(E)<\infty$$. If $$\mu_0$$ is $$\sigma-finite$$, then $$\mu$$ is the unique extension of $$\mu_0$$ to a measure on $$M$$. Image of proof:

$$\textbf{My questions: Does this mean on M, \mu(E)=\mu^{*}(E) for E \subset M?}$$ $$\textbf{My questions: Does this mean on A, \mu(E)=\mu_0(E)=\mu^{*}(E) for E \subset A?}$$

$$\textbf{Does \upsilon, (a measure on M), "extending" \mu_0 mean that the restriction of \upsilon to A is \mu_0 }$$? Folland never explicitly defines extension so I'm not sure what it means.

$$\textbf{Finally, I'm not sure why in the proof why that \upsilon(E) \leq \sum \limits_{1}^{\infty} \mu_0(A_j) }$$, in the screenshot above implies that $$\upsilon(E) \leq \mu(E)$$ Where does this fact come from?

Sorry for all the questions, I just want to really understand everything. Thank you.

• Does this mean on M, $$\mu^*$$ is countably additive? Yes the statement says that $$\mu^*:M\to [0,\infty]$$ is a complete measure. In particular if $$E_n$$ are disjoint sets in $$M$$, then $$\sum_{n=1}^\infty\mu^*(E_n)=\mu^*\left(\bigcup_{n=1}^\infty E_n\right)$$.
• Are all subsets of null sets also contained in M, to be complete? Yes, let $$G\subset A$$, where $$\mu^*(A)=0$$. Since $$\mu^*$$ is an outer measure and $$G\subset A$$, you have that $$0\le \mu^*(G)\le\mu^*(A)=0$$ and so $$\mu^*(G)=0$$. Hence, by what Folland proved you get that $$G\in M$$.
• My questions: Does this mean on M, $$\mu(E)=\mu^{*}(E)$$ for $$E \subset M$$? Actually this is how he defines $$\mu$$. He first defines $$\mu^{*}$$ to be $$\mu^{*}(E):=\left\{\sum \limits_{1}^{\infty} \mu_0(A_j): A_j\in \mathcal{A}, E \subset \bigcup\limits_{i=1}^{\infty}A_j\right\}$$ for every $$E\in X$$. Then he applies Caratheodory's theorem to conclude that $$\mu^*:M\to [0,\infty]$$ is a complete measure. Then he defines $$\mu(E):=\mu^*(E)$$ for every $$E\in M$$. It is just a way for him not to have to write $$\mu^*\vert_M$$ (the restriction of $$\mu^*$$ to $$M$$) every time.
• Does this mean on A, $$\mu(E)=\mu_0(E)=\mu^{*}(E)$$ for $$E \subset \mathcal{A}$$? Yes. I already explained the first equality. Take $$E\in \mathcal{A}$$. Then you can take $$A_1=E$$ and $$A_j=\emptyset$$ for all $$j\ge 2$$ in the definition of $$\mu^{*}(E)$$ to conclude that $$\mu^{*}(E)\le \mu_0(E)+0,$$where we used the fact that $$\mu^{*}(E)$$ is the infimum. To prove the opposite inequality, you have to use the fact that $$\mu_0$$ is a premeasure and $$\mathcal{A}$$ an algebra, then $$\mu_0(E)\le \sum \limits_{1}^{\infty} \mu_0(A_j)$$ for every sequence of sets $$A_j\in \mathcal{A}$$, with $$E \subset \bigcup\limits_{i=1}^{\infty}A_j$$. Try to prove this. Assuming this inequality holds, then $$\mu_0(E)$$ is a lower bound for all such sums $$\sum \limits_{1}^{\infty} \mu_0(A_j)$$ and so the infimum must be greater that $$\mu_0(E)$$, that is, $$\mu_0(E)\le \mu^*(E)$$. Together with the other inequality you have that $$\mu^*(E)=\mu_0(E)$$ for every $$E \subset \mathcal{A}$$.
• Does $$\upsilon$$, (a measure on M), "extending" $$\mu_0$$ mean that the restriction of $$\upsilon$$ to $$\mathcal{A}$$ is $$\mu_0$$? Yes, this is exactly the definition of extension.
• Finally, I'm not sure why in the proof why that $$\upsilon(E) \leq \sum \limits_{1}^{\infty} \mu_0(A_j)$$. Any measure is countably subadditive, that is, if $$E,A_j\in M$$ are such that $$E \subset \bigcup\limits_{i=1}^{\infty}A_j$$, then $$\upsilon(E) \le \sum \limits_{1}^{\infty} \upsilon(A_j).$$ Are you OK proving this property (you have to change the sequence $$A_j$$ to make it disjoint). Now if $$A_j\in \mathcal{A}$$ then you know that $$\upsilon(A_j)=\mu_0(A_j)$$ exactly because $$\upsilon$$ equals $$\mu_0$$ on $$\mathcal{A}$$. So you have $$\upsilon(E) \le \sum \limits_{1}^{\infty} \upsilon(A_j)=\sum \limits_{1}^{\infty}\mu_0(A_j).$$

About the completeness, even with $$\mu^*(G) = 0$$, we still do not know whether $$G \in \mathscr{M}$$ or not. So, the Theorem does not guarantee the completeness of $$\mu$$, and we need to somehow enlarge the domain of $$\mu$$ (e.g., applying Exercise 22 in Folland's book) for completion. Here, remember that $$\mu$$ is the restriction of $$\mu^*: \mathscr{P}(X) \to [0, \infty]$$ on $$\mathscr{M}$$.

To summarize, somehow we have to find a $$\sigma$$-algebra $$\mathscr{\overline M} \supseteq \mathscr{M}$$ that contains every $$G \in \mathscr{P}(X)$$ such that $$\mu^*(G) = 0$$.

We can construct a counter example to show $$\mu$$ is not necessarily complete. Applying the Theorem to a $$\sigma$$-algebra $$\mathscr{A}$$ and any incomplete measure $$\mu_0$$ on it (which are definitely an algebra and a premeasure, respectively). Then,

• the $$\sigma$$-algebra $$\mathscr{M}$$ generated by $$\mathscr{A}$$ is equal to $$\mathscr{A}$$ since $$\mathscr{A}$$ itself is a $$\sigma$$-algebra;

• $$\mu = \mu^*|\mathscr{M} = \mu^*|\mathscr{A}$$, hence by Proposition 1.13 in the Folland's book, $$\mu = \mu_0$$;

• by the construction (i.e., $$\mu = \mu_0$$, $$\mathscr{M} = \mathscr{A}$$), the measure space $$(X, \mathscr{M}, \mu)$$ is the same as $$(X, \mathscr{A}, \mu_0)$$, so not complete.

On the other hand, if $$\mu_0$$ is a $$\sigma$$-finite premeasure, one can construct a $$\sigma$$-finite complete measure by applying the Theorem to obtain a $$\sigma$$-finite measure space $$(X, \mathscr{M}, \mu)$$, where $$\sigma$$-finiteness comes from that of $$\mu_0$$, and then applying Exercise 22 to obtain its completion $$(X, \mathscr{\overline M}, \overline \mu)$$. Such a completion is a unique extension to a complete measure space by Theorem 1.14 above and Theorem 1.9 in the Folland's book.