# Uniqueness of measure on $\mu^*$ measurable sets

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?

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 $$\nu = \mu$$.