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 $$\mu^*(X)=\mu_0(X) ? $$



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, $\mu^{*}(X)=\mu_0(X)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.