Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

$\nu(E) \le \sum_{n\in \mathbb N} \mu_0(F_n)$ 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 $\nu(E)\le \mu(E)$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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