Incomplete measure space that is not sigma-finite 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?
 A: 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.
