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Definition from Folland-Real Analysis,

A function $\mu_0: \mathcal{A} \to [0, \infty]$ will be called a $\textbf{premeasure}$, if

(a). $\mu_0(\emptyset)=0$

(b). if $\{A_j\}_1^{\infty}$ is a sequence of disjoint sets in $\mathcal{A}$ such that $\bigcup_{1}^{\infty} A_j \in \mathcal{A}$, then $$\mu_0 \left(\bigcup_1^{\infty} A_j \right)= \sum_1^{\infty} \mu_0 (A_j)$$

$\textbf{Proposition:}$ If $\mu_0$ is a premeasure on $\mathcal{A}$ and $\mu^*$ is the outer measure induced by $\mu_0$, then

(a). $\mu^* \rvert \mathcal{A}= \mu_0$

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

$\textbf{Question:}$. Let $\mathcal{A} \subset \mathcal{P}(X)$, then since $X \in \mathcal{A}$ and does $u^*$ measurable does the above proposition imply that $$\mu^*(X)=\mu_0(X) ? $$

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Yes.

In the definition of premeasure, Folland requires that $\mathcal{A}$ is an algebra (this is important). So $X\in\mathcal{A}$ by definition. The proposition says that the outer measure $\mu^{*}$ (induced by $\mu_0$) agrees with $\mu_0$ on the algebra $\mathcal{A}$. In particular, $\mu^{*}(X)=\mu_0(X)$.

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