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Folland proves the following theorem:

Let $\mathcal{A} \subset \mathcal{P}(X)$ be an algebra, $\mu_0$ a premeasure on $\mathcal{A}$, and $\mathcal{M}$ be the sigma algbera generated by $\mathcal{A}$. There exists a measure on $\mu$ on $\mathcal{M}$ whose restriction to $\mathcal{A}$ is $\mu_0$ - namely, $\mu = \mu^*|_{\mathcal{M}}$ where $\mu^*$ is the outer measure. If $\nu$ is another measure on $\mathcal{M}$ that extends $\mu_0$, then $\nu(E) \leq \mu(E)$ for all $E \in \mathcal{M}$, with equality when $\mu(E) < \infty$

I have a question about the second part of the statement, showing the equivalence of $\mu$ and $\nu$. When I did this myself, I used the following arguement:

Let $E \subset X$ and $A_j \in \mathcal{A}$, then: $$\nu(E) \leq \sum \nu(A_j) = \sum \mu_0(A_j) = \sum \mu(A_j) = \mu(\bigcup A_j) \, \,\forall j \implies \nu(E) \leq \mu(E)$$ $$\mu(E) \leq \sum \mu(A_j) = \sum \mu_0(A_j) = \sum \nu(A_j) = \nu(\bigcup A_j) \, \,\forall j \implies \mu(E) \leq \nu(E)$$

Folland does this a different way, is this correct though?

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2 Answers 2

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The first assertion follows from Caratheodory's theorem and proposition 1.13 since the $\sigma$-algebra of $\mu^*$-measurable sets include $\mathcal{A}$ and hence $\mathcal{M}$. As for the second assertion, if $E\in \mathcal{M}$ and $E\subset\bigcup_{1}^{\infty}A_j$ where $\{A_j\}_{1}^{\infty}\in\mathcal{A}$, then by montonocity $\nu(E)\leq \nu\left(\bigcup_{1}^{\infty}A_j\right) \leq \sum_{1}^{\infty}\nu(A_j) = \sum_{1}^{\infty}\mu_0(A_j)$, whence $\nu(E)\leq \mu(E)$. If we set $S = \bigcup_{1}^{\infty}A_j$, we have

$$\nu(A) = \lim_{n\to\infty}\nu\left(\bigcup_{1}^{n}A_j\right) = \lim_{n\to\infty}\mu\left(\bigcup_{1}^{n}A_j \right) = \mu(A)$$

If $\mu(E) < \infty$, choose the $A_j$'s such that $\mu(A)\leq \mu(E) + \epsilon$, hence $\mu(A\setminus E) < \epsilon$, and

$$\mu(E)\leq \mu(A) = \nu(A) + \nu(A\setminus E)\leq \nu(E) + \mu(A\setminus E)\leq \nu(E) + \epsilon$$

Since $\epsilon$ is arbitray, $\mu(E) = \nu(E)$. Finally, suppose $X = \bigcup_{1}^{\infty}A_j$ with $\mu_{0}(A_j) < \infty$, where we can assume that the $A_j$'s are disjoint. Then for any $E\in \mathcal{M}$, $$\mu(E) = \sum_{1}^{\infty}\mu(E\cap A_j) = \sum_{1}^{\infty}\nu(E\cap A_j) = \nu(E)$$ so $\nu = \mu$.

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The question here is that one cannot conclude $\mu(E)\leq \nu(E)$ simply by $\mu(E)\leq \nu(\bigcup A_j)$, since $\nu$ is not defined as the restriction of outer measure. In fact, the way you exhibit hasn't used the condition $\mu(E)\leq \infty$

I think the second part can be proved as follows without choosing arbitrary $\epsilon$ as Folland did:

First we assert that, since $\nu(E)\leq\mu (E)\leq\infty$, and by the definition of $\mu$, one can find some $A=\bigcup_{A_j\in \mathcal A} A_j\supset E$ s.t. $\mu(A)\leq\infty$, and $\mu(A)=\nu(A)$. Hence$$\mu(A)-\mu(E)=\nu(A)-\mu(E)\leq\nu(A)-\nu(E)=\nu(A-E)\leq\mu(A-E)\leq\infty$$
This implies $\mu(E)=\nu(E)$.

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