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When defining what “a measurable set” means, one quite naturally arrives at the definition of an algebra:

  1. $\emptyset$ and $X$ are measurable
  2. if $A$ is measurable, then $X \setminus A$ is measurable
  3. if $A$ and $B$ are measurable, then $A \cup B$ is measurable

Each of these rules makes perfect sense to me. However, the actual definition of measurability uses the rules of a $\sigma$-algebra, adding the union of an countably infinite number of sets:

  1. if $A_i$ are measurable, where $i \in \mathbb N$, then $\bigcup_i A_i$ is measurable

Why do we need this stricter definition? I know the definition of the Lebesgue integral uses the approximation by simple functions with countably many steps, but it isn't clear to me at all that this wouldn't also work with finite partitions.

Is there a simple and intuitive reason, why we must require the countable union? If not, is there a subtle, technical reason for it? Or do we require it simply because it is more convenient to work with?

There have been questions which touched upon this topic [1] [2] [3], but as far as I can tell, nobody has provided a direct satisfactory answer to this question.

csha
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    How would you deal with showing that, say, the cantor set is measurable with only finite steps allowed? – Alan Aug 13 '21 at 21:17
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    Another aspect may come from taking the products. If we only allow finite unions of intervals on the line, then we only get finite unions of rectangles on the plane. But we also want to measure, say, a triangle or a disk. Both with grade school formulas for the area. – Jyrki Lahtonen Aug 13 '21 at 21:22
  • related: https://math.stackexchange.com/questions/4070946/integration-with-respect-to-a-finitely-additive-measure – Alan Aug 13 '21 at 21:34

2 Answers2

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As Mark Bennet's answer states, the reason is most likely that we want to be able to approximate sets (e.g. approximate a circle with rectangles and get a guarantee that the limit is sensible)

While this property is certainly useful, it might be too greedy to require a measure to have it:

It is well known that the Maximum a posteriori estimator coincides with the maximum liklihood if the prior is uniform. But on the real line a uniform distribution does not exist. But the concept of an "uninformative prior" is so useful to Bayesian statisticians, that they try to make this work as an "improper prior". In the book Theory of statistics, page 21, Mark Schervish mentions that a possible solution to this problem due to DeFinetti (1974) is to relax the requirement that measures need to be countably additive. I unfortunately never had the time to read the original source (most likely referring to "Theory of Proability: Volume I", which was compressed together with volume II into one book "Theory of Probability: A critical introductory treatment" in 2017.

A theory of finitely additive measures

In the book Infinite Dimensional Analysis (Aliprantis and Border 2006), the concept of "Charges", i.e. finitely additive measures is analysed. They state in the introduction of Chapter 12

Indeed if a charge (which need only be finitely additive) on the Borel sets of a Hausdorff space is tight, then it is automatically countably additive (Theorem 12.4).

That is, tight finitely additive "measures" are sigma additive measures. So perhaps this restriction of sigma additivity limits measures to this special case.

Disclaimer: I have no idea of the ramifications of removing this assumption as I never had the time to research these in detail but the references provided should be helpful to someone interested in this topic equipped with more time.

Felix Benning
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Perhaps that measure wants to capture limiting processes, and therefore ought to have an infinite aspect?

If you don't have countable unions, you essentially add nothing to what is already known?

There are two suggestions to begin ...

Mark Bennet
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