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By definition we only allow the union of countable infinite of elements to be also include the $\sigma$ field, why not uncountable many? Is there a historical view behind this?

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If we require sigma algebras to be closed under arbitrary unions, then any sigma algebra on $\mathbb{R}$ that contained the open intervals would necessarily contain every subset of $\mathbb{R}$.

In particular, things like Vitali sets would have to be measurable.

As a consequence, any translation-invariant measure on $\mathbb{R}$ would have to either (a) be the zero metric or (b) assign infinite measure to finite intervals.

This would make measure theory kind of useless.

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  • $\begingroup$ Thanks! I see your point. Could you please elaborate your definition for "closed under uncountable union?" What would be an example that is not closed under uncountable union? $\endgroup$ Jan 23, 2016 at 19:55
  • $\begingroup$ A sigma algebra on a set $X$ is, by definition, a subset $\Sigma \subseteq \mathcal{P}(X)$ which is closed under countable unions and intersections. The question is about changing that definition to "closed under arbitrary unions and countable intersections." So essentially every sigma algebra commonly used would be an example of a sigma-algebra that's not closed under uncountable unions. $\endgroup$ Jan 23, 2016 at 19:59
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    $\begingroup$ @KennethChen Well, the Borel sigma-algebra on the unit interval is not closed under uncountable union, as has been stressed several times on the page already. $\endgroup$
    – Did
    Jan 23, 2016 at 20:00
  • $\begingroup$ @Did Yes, I got it! $\endgroup$ Jan 23, 2016 at 20:01
  • $\begingroup$ @DanielMcLaury I got it! Thanks a lot for your kind help and time! $\endgroup$ Jan 23, 2016 at 20:01
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The borel sigma algebra, which is used in many cases, contains every single point of the underlying space as an element. If you'd allow uncountable unions, the borel sigma algebra would have to contain any subset of the space.

However lots of measures cannot be constructed in a way that every set is measurable, a very important example of this is the measure on $\mathbb{R}$ that maps each interval to its length. There is a proof that it cannot be constructed such that every set is measurable.

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  • $\begingroup$ To be a little bit more explicit: he means the Lebesgue measure. $\endgroup$
    – noctusraid
    Jan 23, 2016 at 20:18
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You can show, that it's not possible to construct a measure $\lambda$ defined on all the subsets of $\mathbb{R}$ such that the following natural assumptions hold:

  1. For intervals $[a,b]$, you have $\lambda([a,b])=b-a$.
  2. Translation Invariance: $\forall X\subset\mathbb{R}$ and $\forall t\in\mathbb{R}$ you have $$\lambda(X+t)=\lambda(\{x+t | x\in X\})=\lambda(X)$$
  3. Additivity: For any sequence $(X_i)_{i\in I}$ of disjoint sets in $\mathbb{R}$ you have $$\lambda\Bigl(\bigcup_{i\in I} X_i \Bigr)=\sum_{i \in I}\lambda(X_i)$$

And therefore you need to have some restrictions on your sets (at least if you want to keep the assumptions about your measure, which seems like a good idea)

Counter example (vitali sets):

Define the following equivalence relation on $\mathbb{R}$: $$x \sim y \quad \text{iff} \quad x-y \in \mathbb{Q}$$

Then, by the axiom of choice, there is a map $f\colon \mathbb{R} / \sim \to \mathbb{R}$ which assigns a representant to each equivalence class. You can assume that $X=f(\mathbb{R} / \sim)\subset [0,1]$ (you can just "add integers").

Now look at $\lambda(X)$:

  1. If $\lambda(X)$ would be equal to $0$, you would have $$\lambda(\mathbb{R})=\lambda \Bigl(\bigcup_{\mathbb{Q}}(t+X) \Bigr)=\sum_{\mathbb{Q}}\lambda(X)=0$$ by the translation invariance and the countability of $\mathbb{Q}$.
  2. If $\lambda(X)=c>0$, you get $$\lambda \Bigl(\bigcup_{\mathbb{Q}\cap [0,1]}(t+X) \Bigr)=\sum_{\mathbb{Q}\cap [0,1]} c = +\infty$$ But you also have $$\lambda \Bigl(\bigcup_{\mathbb{Q}\cap [0,1]}(t+X) \Bigr)\leq \lambda([0,2])=2$$ and therefore another contradiction.
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