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I am reading Folland's Real analysis book and I saw following sentence in the Theorem 1.14:

Let $A \subset \mathcal{P}(X)$ be an algebra, $\mu_0$ a premeasure on $\mathcal{A}$, and $\mathcal{M}$ the $\sigma$-algebra generated by $\mathcal{A}$. There exists a measure $\mu$ on $\mathcal{M}$ whose restriction to $\mathcal{A}$ is $\mu_0$ -- namely, $\mu = \mu^* | \mathcal{M}$ (where $\mu^*$ is outer measure).

What I want to ask is, what book means by saying There exists a measure $\mu$ on $\mathcal{M}$ whose restriction to $\mathcal{A}$ is $\mu_0$?

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2 Answers 2

up vote 2 down vote accepted

There exists a measure $\mu$ on $\mathcal M$ that gives you the same value for each element of $\mathcal A$ as $\mu_0$ would give you. That's all.

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$\mu$ gives the same value fore each element of $\mathcal A$, not subset. –  tomasz Nov 26 '12 at 22:40
    
yes, of course .. i've edited my answer. ty –  user127.0.0.1 Nov 26 '12 at 22:44

This means that if you define $ \mu(E) = \mu^*( E) $ for all $ E \in \mathcal{M} $, then $ \mu $ is a measure on $\mathcal{M}$.

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