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I saw definition of complete measure space as a set of subsets of X with a measure defined on it so that if a set is null(with measure zero) then all its subsets are in algebra(and then obviously have measure zero). Completion in what sense? And completion of measure space is unique. Does it mean if I take any superset of completion of measure space, then it is not measure space i.e. sigma additivity does not hold in it.

Sushil
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1 Answers1

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You should look at Exercises 1.2.23 and 1.2.24 from Terry Tao's measure theory text. Even if you don't do them, they give a good sense of how completion of a measure relates to the completion of an associated (pseudo)metric space. So completeness can be taken in this sense, but this might not be the best way of thinking about it.

Alternatively, you can view a completion as being the smallest extension of a measure which allows you to measure all subsets of null sets. In this sense, the completion of a measure is unique - i.e. there is only one smallest extension, and it's formed by just taking the $\sigma$-algebra generated by your original measurable sets plus the sub-null sets and deciding that the sub-null sets have measure zero.

To answer your last question: no, it doesn't mean that any larger $\sigma$-algebra fails to admit an extension of your original measure. For example, if $X = \{1, 2, 3\}$, $\chi= \{\emptyset, \{1\}, \{2, 3\}, \{1,2,3\}\}$, and $\mu$ counts the number of elements in a set, then the completion of $(X, \chi, \mu)$ is exactly $(X, \chi, \mu)$ (because $\mu$ was already complete). But we could take the discrete $\sigma$-algebra on $X$ with counting measure, and this would be a complete extension of $\mu$. The completion of a measure is only unique because we define it to be the $\textbf{smallest}$ extension that lets us measure sub-null sets.

Josh Keneda
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  • You gave an example of an already complete set $\chi$, can you give an example of a non-complete set, and then what to add to make it complete ? – matthieu Feb 28 '18 at 09:53
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    I would guess that the most important example of something non-complete would be the Borel $\sigma$-algebra on $\mathbb{R}$. The Borel $\sigma$-algebra has cardinality $\frak c$ (the same cardinality as the reals), but its completion would have to contain all subsets of a null Cantor set (for example), so the completion would have cardinality $2^{\frak c}$, which is strictly bigger. So the Borel $\sigma$-algebra is too small to be complete. Tossing in all Lebesgue-null sets fixes the problem. – Josh Keneda Feb 28 '18 at 21:16