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Let $(X,\mathcal A_0,\mu_0)$ be a measure space with $\mathcal A_0$ being an algebra. We can always extend $\mu_0$ by Carathéodory's method to $(X,\mathcal A,\mu)$, with $\mathcal A\supset \mathcal A_0$ a $\sigma$-algebra and $\mu\equiv\mu_0$ in $\mathcal A_0$.

Hahn's Extension Theorem states that if $\mu_0$ is $\sigma$-finite, any method other than Carathéodory's will generate the same measure space $(X,\mathcal A,\mu)$. Here's how Bartle's The Elements of Integration and Lebesgue Measure proves it (1st edition, p. 103) when $\mu_0$ is finite.


Let $\nu$ be a measure that coincides with $\mu_0$ in $\mathcal A_0$. Then $\mu$ and $\nu$ are finite and coincide in $\mathcal A_0$. Let's take $E\in\mathcal A$ and $\langle F_n\rangle_{n\in\mathbb N}\subset\mathcal A_0$ such that $E\subset\bigcup_{n\in\mathbb N}\;F_n$; so $$ \nu(E)\le\nu\left(\bigcup_{n\in\mathbb N}\;F_n\right)\le\sum_{n\in\mathbb N}\;\nu(F_n)=\sum_{n\in\mathbb N}\;\mu_0(F_n)\quad,$$ thus $\nu(E)\le\mu(E)$. Since $X\in\mathcal A_0$, $$ \mu(X)=\nu(X)\quad\therefore\quad\mu(E)+\mu(E^\complement)=\nu(E)+\nu(E^\complement)\quad.$$ Using the finitess of these measures, we obtain $$ \mu(E)-\nu(E)=-\,\mu(E^\complement)+\nu(E^\complement)\ge0\quad\therefore\quad\nu(E^\complement)\ge\mu(E^\complement)\quad.$$ As $E$ is arbitrary, $\nu(E)\ge\mu(E)$, and we conclude that $\mu\equiv\nu$ in $\mathcal A$.


I see one problem here: as usual, Bartle proves thing too fast, and I don't know how he got $\nu\le\mu$ in $\mathcal A$. Why does the inequality involving $E$ and $\bigcup_{n\in\mathbb N}\;F_n$ imply that?

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To understand this statement you need to remember the definition of outer measure ($\mu_0^*$) and remember that $\mu_0^*(E)=\mu(E)$ if $E$ is a measurable with respect to $\mu_0^*$. –  Leandro Dec 22 '11 at 8:15
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$\nu(E) \le \sum_{n\in \mathbb N} \mu_0(F_n)$ tells you that $\nu(E)$ is a lower bound on $\{\sum_{n\in \mathbb N} \mu_0(F_n) | E\subset \cup_{n\in \mathbb N}F_n\}$. By definition $\mu(E) = \inf\{\sum_{n\in \mathbb N} \mu_0(F_n) | E\subset \cup_{n\in \mathbb N}F_n\}$ and so $\mu(E)$ is also a lower bound on $\{\sum_{n\in \mathbb N} \mu_0(F_n) | E\subset \cup_{n\in \mathbb N}F_n\}$ but its the greatest lower bound on it. Therefore $\nu(E)\le \mu(E)$.

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Ooooh, so it was an infimum argument! Like I said, Bartle goes way too fast. –  Luke Dec 22 '11 at 20:25
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