I'm reading "Real Analysis" by Royden (4th ed). According to this text, a semiring $ S$ of subsets of $X$ is a nonempty collection of subsets of $X$ that is closed for finite intersections and such that, if $A,B \in S$, then $A\setminus B = \bigcup\limits_{k=1}^{n} C_{k}$, with the $C_k \in S$ and disjoint. Given a collection $S$ of subsets of $X$, a premeasure $\mu$ is a map $\mu :S \rightarrow \left [ 0, \infty \right ] $ for which $\mu\left (\bigcup\limits_{k=1}^{n} E_{k} \right ) = \sum \limits_{k=1}^{n} \mu\left ( E_k \right )$ whenever the $E_k \in S$ are disjoint and $\bigcup\limits_{k=1}^{n} E_{k} \in S$ and $\mu\left ( E \right ) \leqslant \sum \limits_{k=1}^{\infty} \mu\left ( E_k \right )$ whenever $E, E_k \in S$ with $ E \subseteq \bigcup\limits_{k=1}^{n} E_{k} $ and for which $\mu\left(\emptyset\right ) = 0$ if $\emptyset \in S$ . Then, the following theorem gets proven (I get all that):

Let $\mu :S \rightarrow \left [ 0, \infty \right ] $ be a premeasure on a semiring $S$ of subsets of $X$. Then the Carathéodory measure $\bar{\mu}$ induced by $\mu$ is an extension of $\mu$. Furthermore, if $\mu$ is $\sigma$-finite, then $\bar{\mu}$ is the unique measure on the $\sigma$-algebra of Carathéodory measurable sets that extends $\mu$.

What I don't understand is why the following is a corollary of this theorem: Let $S$ be a semiring of subsets of $X$ and $\mathcal{B}$ the smallest $\sigma$-algebra of subsets of $X$ that contains $S$. Then two $\sigma$-finite measures on $\mathcal{B}$ are equal if and only if they agree on $S$. What am I missing? Any thoughts anyone? Many thanks in advance.


The corollary is indeed not that straightforward. It could be conceivable that there are two different extensions to the $\sigma$-algebra generated, but only one extends to the $\sigma$-algebra of of all Caratheodry-measurable sets. So one has to use a little bit more.

Here is how one can do it: Let $\mu_1$ and $\mu_2$ be two $\sigma$-finite measures on the smallest $\sigma$-algebra containing all sets in $S$, which we denote by $\sigma(S)$. We can extend both $\sigma$ algebras to the $\sigma$-algebras of the respective $\sigma$-algebras of Caratheodory-measurable sets $\mathcal{C}_1$ and $\mathcal{C}_2$. Let $\mathcal{C}$ be the $\sigma$-algebra of Caratheodory measurable sets one gets from starting with $\mu$ and $S$. Now here is the fact one has to use: We have both $\mathcal{C}\subseteq\mathcal{C}_1$ and $\mathcal{C}\subseteq\mathcal{C}_2$. So restricting $\bar{\mu}_1$ and $\bar{\mu}_2$ to $\mathcal{C}$, we get two measures on $\mathcal{C}$ that extend $\mu$. By the uniqueness of extension, these restrictions coincide. So the restrictions to the even smaller $\sigma$-algebra $\sigma(S)$ have to coincide too. But these restrictions are just $\mu_1$ and $\mu_2$. So $\mu_1=\mu_2$.

  • $\begingroup$ I suppose that by μ you mean the restriction of μ1 and μ2 to S. But, it is not that obvious (but quite possible, I don't have a clue) that one can apply the uniqueness of extension, because the semi-finiteness is not given for μ but for μ1 and μ2. $\endgroup$ – user58664 Jun 21 '13 at 11:37
  • $\begingroup$ @user58664 I'll think about that point, thanks. $\endgroup$ – Michael Greinecker Jun 21 '13 at 12:10

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