Given a $\sigma$-ring.

Consider a premeasure on it: $$\mu:\mathcal{R}\to\overline{\mathbb{R}}_+:\quad\mu(R)=\sum_{k=1}^\infty\mu(R_k)\quad\left(R=\biguplus_{k=1}^\infty R_k\right)$$

Extend this to a measure by: $$\mu_E:\sigma(\mathcal{R})\to\overline{\mathbb{R}}_+:\quad\mu_E(A):=\infty\quad\left(A\in\sigma(\mathcal{R})\setminus\mathcal{R}\right)$$

Does it coincide with Caratheodory's extension?

(Caution, the extension may not be unique!)

  • $\begingroup$ Besides, does someone know how I can implement a disjoint union symbol like \bigcupdot or \bigcupplus in mathjax? $\endgroup$ Feb 17, 2015 at 19:35
  • 1
    $\begingroup$ Are you searching for $\biguplus$? (This is done with "\biguplus"). $\endgroup$
    – PhoemueX
    Feb 17, 2015 at 22:56
  • $\begingroup$ @PhoemueX but what about $\bigcupdot$ $\endgroup$
    – Alec Teal
    Mar 13, 2015 at 19:32

1 Answer 1


In case of $X \in \mathcal{R}$, $\mathcal{R}$ is actually a $\sigma$-algebra, so that the claim is trivially true.

Now assume that $A \in \sigma(\mathcal{R}) \setminus \mathcal{R}$. By this question (Generating a $\sigma$-algebra from a $\sigma$-ring) this means $A^c \in \mathcal{R}$.

Now, the Caratheodory extension must give $\mu^\ast(A) = \infty = \mu_E (A)$, because otherwise there would be a countable covering $A \subset \bigcup_n R_n$ of $A$ with elements of $\mathcal{R}$ (even with $\sum_n \mu(R_n) < \infty$, but we do not need that).

But this implies $X = A \cup A^c = A^c \cup \bigcup_n R_n \in \mathcal{R}$, in contradiction to our assumption.

  • $\begingroup$ Ah right. Btw thanks for the link! $\endgroup$ Feb 17, 2015 at 23:54

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