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A measure space $(X,\mathfrak{M},\mu)$ is decomposable if $X$ is a disjoint union of measurable subsets, $X=\bigcup_{i\in I}X_{i}$, with $\mu(X_{i})<\infty$ for all $i$, and $\mu(A)=\sum_{i\in I}\mu(A\cap X_i)$ for every measurable set $A$ of finite measure.

It is easy to see that a $\sigma$-finite measure space is decomposable.

Question: What is an example of a decomposable measure space that is not $\sigma$-finite?

A `simplest possible' example would be most welcome.

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up vote 4 down vote accepted

It is clear that every countable index set will be insufficient.

Therefore, it is natural to look at $\Bbb R$ as an index set; consider it as a disjoint set like so:

$\displaystyle \quad \Bbb R = \bigcup_{x \in \Bbb R} \{x\}$

Then put on this the discrete $\sigma$-algebra $\Sigma$, and put on $(\Bbb R, \Sigma)$ the counting measure $\mu$. Obviously, only countably many $\{x\}$ can be attained by an increasing sequence of sets of finite measure.

That is, $(\Bbb R, \Sigma, \mu)$ is not $\sigma$-finite.

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Thanks! This was as trivial as I thought it might be... – Jon Bannon Oct 11 '12 at 15:43
You're welcome; sometimes a fresh look at the problem is all one needs. – Lord_Farin Oct 11 '12 at 16:59

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