I am looking for an example of an incomplete measure space with a measure that is not sigma-finite.

All the measures which are not sigma-finite which I have come across so far are the following:

  • counting measure on a set that is not countable (e.g. on the measurable space $(\mathbb{R},\mathcal{P}(\mathbb{R}))$
  • or the measure $\mu$ on the trivial sigma algebra $\Sigma = \{\emptyset,X \}$ with $\mu(X) = \infty$

were complete.

Suppose $(\Omega, \Sigma, \mu)$ would be such a measure space. Let $\eta$ denote the induced outer measure and $\Sigma_\eta$ the $\sigma$-algebra of the $\eta$-measurable sets. Furthermore let $(\Omega,\tilde{\Sigma},\tilde{\mu})$ denote the completion of $(\Omega,\Sigma,\mu)$. Will in this case the inclusion $\tilde{\Sigma} \subset \Sigma_\eta$ be strict?

  • 1
    $\begingroup$ Take an uncountable direct sum of $([0,1], {\rm Borel}([0,1]), \lambda)$ where $\lambda$ is the Lebesgue measure. $\endgroup$ Aug 3, 2015 at 22:09
  • $\begingroup$ Another example would be the long line with its natural Borel measure structure. en.wikipedia.org/wiki/Long_line_%28topology%29 $\endgroup$
    – user24142
    Aug 3, 2015 at 22:18

1 Answer 1


A very simple example: \begin{align*} X\equiv&\,\{a,b,c\},\\ \Sigma\equiv&\,\{\varnothing,\{a,b\},\{c\},\{a,b,c\}\}, \end{align*} with \begin{align*} \mu(\varnothing)\equiv&\,0,\\ \mu(\{a,b\})\equiv&\,0,\\ \mu(\{c\})\equiv&\,\infty,\\ \mu(\{a,b,c\})\equiv&\,\infty. \end{align*} Then, $\Sigma$ is a $\sigma$-algebra, and $\mu$ is a measure on it. However, $\mu$ is

  • not $\sigma$-finite: no measurable set containing $c$ has finite measure; and
  • not complete: $\{a\}\subset\{a,b\}$ and $\mu(\{a,b\})=0$, but $\{a\}\notin\Sigma$.

The induced outer measure $$\eta(A)\equiv\inf\left\{\sum_{n=1}^{\infty}\mu(S_n)\,\Bigg|\,S_1,S_2,\ldots\in\Sigma\text{ and }A\subseteq\bigcup_{n=1}^{\infty}S_n\right\}\quad\forall A\subseteq X$$ is as follows: \begin{align*} \begin{array}{rclcrclcrclcrcl} \eta(\varnothing)&=&0&&\eta(\{a\})&=&0&&\eta(\{a,b\})&=&0&&\eta(\{a,b,c\})&=&\infty\\ &&&&\eta(\{b\})&=&0&&\eta(\{a,c\})&=&\infty&&&&\\ &&&&\eta(\{c\})&=&\infty&&\eta(\{b,c\})&=&\infty \end{array} \end{align*} It can be shown that any subset of $X$ is $\eta$-measurable, so $\Sigma_{\eta}=2^X$. In fact, $2^X$ is also the completion $\tilde{\Sigma}$ of $\Sigma$ with respect to $\mu$. Therefore, $\tilde\Sigma=\Sigma_{\eta}$.

I don't know if, in general, one can have $\tilde{\Sigma}\subset\Sigma_{\eta}$ strictly.


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