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My lecture notes on measure theory motivate countable additivity of measures by the method of exhaustion, which makes perfect sense to me.

After this the notes pose the question whether we could also demand uncountable additivity and then state the following theorem.

Theorem: Let $I$ be a set and $a: I \to (0,\infty)$ a map such that $\sum \limits_{k \in I} a_k$ converges. Then $I$ is countable.

(Please note that the convergence of an uncountable sum can be defined in terms of nets (see here).)

The notes then mention that this means we only need countable additivity.

I understand the theorem and that this means that uncountable additivity of measures would mean that any uncountable union of measures would have infinity measure. But why is this a problem? Does it break anything? One thing I came up with as an explanation is that the measure of an uncountable sum does not really depend on the exact measures of the single sets. As long as uncountable many sets have positive measure the union will have infinite measure (unlike for a countable sum).

Thanks for any help and suggestions!

Edit: I know that in the end uncountable additivity would imply that every subset of $\mathbb{R}$ has Lebesgue measure $0$, but I feel like the author of the notes has something else in mind. In fact, at this stage we don't even know the Lebesgue measure.

DerivativesGuy
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    What happens to Lebesgue measure? $m(A)=\sum_{x \in A} m({x})=\sum 0=0$ for every $A$! – Kavi Rama Murthy May 30 '22 at 06:09
  • @KaviRamaMurthy Yes, I know this argument, but I think this is not what the author of the lectures notes is getting at. In fact, the author mentions the theorem above at the very start and we don't know that single points have measure $0$ yet. – DerivativesGuy May 30 '22 at 06:22
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    There are measures which are uncountably additive. $\mu (\emptyset)=0, \mu (A)=\infty$ if $A$ is non-emtpy defines an uncountably additive measure on any set. So uncountable additivity does not lead to any contradiction. – Kavi Rama Murthy May 30 '22 at 06:28

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Say $\mu$ is an uncountably additive measure on some set $X$ and $\mathcal{S}\subseteq \mathcal{P}(X)$ is an uncountable collection of measurable sets. Then since expressions such as: $$\mu\left({\bigcup \mathcal{S}}\right)=\sum_{A\in \mathcal{S}}\mu\left(A\right)$$ would now be meaningful, using a $\sigma$-algebra $\mathfrak{m}$ to represent the domain of $\mu$ is no longer sufficient, as we have no guarantee that ${\bigcup \mathcal{S}}\in\mathfrak{m}$.

If we try to extend $\mathfrak{m}$ so as to make equalities such as the above meaningful, we'd be unable to develop any meaningful measure that naturally extends the concepts of length/area/volume, since sets that cannot be measurable would be considered measurable with these definitions.

SomeStrangeUser
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  • Thanks for your answer. I get the argument, but actually the notes mention uncountable additivity even before stating the fact that it is impossible to measure all sets if we allow the axiom of choice. So I am looking for a more intuitive reason why the fact that uncountable sums always have infinite sum suggests that uncountable additivity is a bad idea. Countable sums are not always infinite, so it feels like the notes sort of say that you get too many sets with measure infinity or something. Also if we change one set of an uncountable union the measure of the union won't ever change. – DerivativesGuy May 30 '22 at 09:18
  • I've given this another thought. I think we can argue like this instead: If I want to assign measures to all sets (Let's pretend that I don't know that this is impossible yet.), then a single point $x \in \mathbb{R}$ either has measure $0$ or positive measure. But with uncountable additivity this implies that all uncountable sets would have to either have measure $0$ or $\infty$. In partcular, $[a,b]$ would not have measure equal to $b-a$. Hence, with this notion there is no extension of length to a measure with the properties we want. – DerivativesGuy May 31 '22 at 06:42
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    @DerivativesGuy My initial answer included that argument, but I felt it was too close to (an immediate consequence of) the argument already made in the comments to your question ($\sum_{x\in A} m({x})=0$), so I edited it out. – SomeStrangeUser May 31 '22 at 15:28