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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:

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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: enter image description here

$\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.

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Your questions:

  • 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 $$\upsilon(E) \le \sum \limits_{1}^{\infty} \upsilon(A_j)=\sum \limits_{1}^{\infty}\mu_0(A_j). $$
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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.

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