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I was just wondering why one always insists on countability when it comes to the definition of a $\sigma$-algebra in measure theory. I mean, measure theory works as it does, but is there a deeper reason why e.g. it is reasonable to demand that the countable union of measurable sets is measurable again? Why not an uncountable union? Thanks in advance!

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    $\begingroup$ Because everything is a (possibly) uncountable union! $\endgroup$ – Schemer Sep 3 '15 at 22:23
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    $\begingroup$ One should do the exercise of proving that a convergent sum of positive real numbers must be countable. $\endgroup$ – paul garrett Sep 3 '15 at 22:41
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    $\begingroup$ You could look at Chapter 1 of these notes terrytao.files.wordpress.com/2011/01/measure-book1.pdf which refer helpfully to the Banach-Tarski paradox, the question of the measure of the intervals $[0,1]$ and $[0,2]$ which can easily be related by a bijection, and the Axiom of Choice, which lies subtly behind the answer to your question. $\endgroup$ – Mark Bennet Sep 3 '15 at 22:53
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    $\begingroup$ For the other side of the question -- why countable unions and not just finite? -- it turns out that in three dimensions (and up) we seem to need limiting processes to give a decent theory of volume even for polyhedra. See Hilbert's 3rd problem. $\endgroup$ – user21467 Sep 4 '15 at 0:03
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If you insist on uncountable unions, then the measure of the unit interval is the sum of uncountably many measures of the individual points of the interval. These measures are either 0 (in which case $\mu(I) = 0$, which is bad) or nonzero, in which case $\mu(I)$ would be $\infty$,.

A possibly more interesting question is whether finitely-additive measures would be interesting. Bumby and Ellentuck wrote an interesting (to me) paper on the topic, although perhaps part of the interest was that it was the first "real math research paper" I ever read. :)

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It's quite simple: Allowing only finite union yields too weak results [cf. Jordan measure] and allowing uncountable unions makes the system of measurable sets too big. If you for example assume that all singletons are measurable, then uncountable unions would imply that any set is measurable. This already doesn't work well with the Lebesgue measure, as the example of Vitali sets show.

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Another reason is that in a summable family $(x_i)_{i\in I}$, it can be shown the set $\;\{i\in I\mid x_i\neq 0\}$ is at most countable. Thus the theory of summable families comes down to the theory of series.

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Countable has this very peculiar property of infinity.

It can be indexed, such that every point is only a finite distance away. Namely, it can be indexed using $\Bbb N$. And $\Bbb N$ has that great property that given any $k\in\Bbb N$, you only need to travel finitely many steps to meet with $k$.

Why is that important? Because (unfortunately) mathematics and mathematicians take their intuition from the real world. This means that you develop probability and measure from finite experiments. Often finite experiments have "unimportant residue", which is to say that the more times you run something, the closer that residue gets to $0$. So when you take a limit, you get this squeaky clean result.

Well... sometimes, anyway.

But this is where countable unions come from. First we cared about running an experiment $n$ times, then we also noticed that when $n\to\infty$ things look nicer. But what does it mean $n\to\infty$ when running an experiment? It means that we have a countable sequence of events, and we want to know whether the limit of those events is an events. We sure want it to be. So why not make it? And so $\sigma$-algebras are born.

(All that being said, there are uses - interesting uses - to algebras which have closure of more than just countable unions. But that usually gets under set theory, rather than measure theory or probability.)

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  • $\begingroup$ Would another way to phrase this answer (sort of) be: "Countability allows induction"? I'm getting that by extrapolating from "we also noiticed that when $n\to\infty$ things look nicer." If that's not part of what you're talking about, it seems like it is part of why we would want to know/define when things are countable. $\endgroup$ – Todd Wilcox Sep 4 '15 at 11:59
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    $\begingroup$ Todd, that's not entirely correct since well-orders allow induction. And there are uncountable well-orders. The thing I meant about things being nicer at the limit is that often you have something like $x^2+o(x^3)$ which will limit at $x^2$. So while finite things require you to calculate some explicit expression (for better or worse), limits allow you to discard that which does not matter. $\endgroup$ – Asaf Karagila Sep 4 '15 at 12:21
  • $\begingroup$ (The $x^2+o(x^3)$ is probably meaningless, but you get my gist.) $\endgroup$ – Asaf Karagila Sep 4 '15 at 12:22
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I hope this works (check this): Counterexample why it is a bad idea to consider uncountable unions:

Take $\lambda \in [0,1] \setminus \mathbb{Q}$ and define $A_\lambda := \{\lambda + r \,|\,r\in\mathbb{Q}\} \cap [0,1]$. Then the sets $A_\lambda$ are pairwise disjoint. Now we have

$$\mu(\bigcup_{\lambda \in [0,1] \setminus \mathbb{Q}} A_\lambda) = \mu([0,1]) \neq \sum_{\lambda \in [0,1] \setminus \mathbb{Q}} \mu(A_\lambda) = 0.$$

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    $\begingroup$ Why don't you just take $A_\lambda = \{\lambda\}$ and take the union over all $\lambda \in [0,1]$? $\endgroup$ – Dominik Sep 4 '15 at 7:49
  • $\begingroup$ @Dominik yes, that would work of course $\endgroup$ – Loreno Heer Sep 4 '15 at 9:13

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