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Problem. Let $X$ be a set, and define a function $\mu : \mathcal{P}(X) \to [0, \infty]$ by

$$ \mu(A) = \begin{cases} \infty & \text{if } A \text{ is uncountable} \\ 0 & \text{otherwise} \end{cases} $$

Is $\mu$ $\sigma$-additive?

Attempt. I know that $\sigma$-additivity means that for a countable index set $I$, the following must hold:

$$\mu \left(\bigcup_{i \in I} A_i\right) = \sum_{i \in I} \mu (A_i) \quad when \quad A_i \cap A_j = \emptyset \quad for \quad i \neq j$$

If $\forall i \in I : A_i$ are countable, then $\mu(A_i) = 0$, and since the countable union of countable sets is countable, we have:

$$\mu\left(\bigcup_{i \in I} A_i\right) = 0 = \sum_{i \in I} \mu(A_i)$$

If at least one $A_i$ is uncountable, then $\mu(A_i) = \infty$ for at least one index $i$, so:

$$\sum_{i \in I} \mu(A_i) = \infty$$ The union will then also be uncountable, because $ A_i \subset \cup A_i $.

Question. I am not sure whether my argument is correct, or if I might be completely off track. I would greatly appreciate any help, feedback, or suggestions to better understand whether this function is actually $\sigma$-additive and how to approach this properly. Thank you very much!

j.primus
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    Your argument looks correct to me. Only error I can see is in the first statement; $\sigma$-additivity only says that $\mu(\bigcup A_i) = \sum \mu(A_i)$ holds when the $A_i$ are disjoint. However, this does not really affect the argument. In fact, for this particular measure, $\mu(\bigcup A_i) = \sum \mu(A_i)$ holds even when the sets are not disjoint, as your argument shows. – ummg Mar 07 '25 at 18:03
  • @ummg Thank you, you're absolutely right — I completely forgot about that. – j.primus Mar 07 '25 at 18:13
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    The answer depends on your axiom system for sets. The point is that the satement "the countable union of countable sets is countable" isn't provable in ZF and there are models of set theory where even the countable union of disjoint two-element sets isn't countable. – tj_ Mar 07 '25 at 19:49
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    @tj_ Sure, but if you don't have at least countable choice, there's not much point in trying to do measure theory! It's safe to assume that questions about measure theory are working in the context of full choice (and more generally, it's safe to assume ZFC foundations unless otherwise specified). – Alex Kruckman Mar 07 '25 at 22:28
  • Yes, your argument is correct. – Ramiro Mar 08 '25 at 00:40

1 Answers1

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Your argument's correct – but it actually generalizes quite a bit, and I think it's worth mentioning! Let $X$ be a set and $I\subseteq \mathcal P(X)$. We'll say $I$ is a $\sigma$-ideal when:

  1. $\varnothing \in I$
  2. $I$ is closed under subsets, i.e. $B \in I$ and $A\subseteq B$ implies $A\in I$,
  3. $I$ is closed under countable unions.

For example, $\{A\subseteq X \mid A \text{ countable}\}$ is a $\sigma$-ideal. The prototypical example of a $\sigma$-ideal is the nullsets of a measure $\mu: \mathcal P(X)\to [0,\infty]$. Indeed, $\mu(\varnothing) = 0$, by monotonicity subsets of nullsets are null, and similarly countable unions of nullsets are null. (If you haven't proven this before, consider it a useful exercise.)

Your argument, abstracted a bit, shows that every $\sigma$-ideal $I\subseteq \mathcal P(X)$ arises as the nullsets of a measure $\mu: \mathcal P(X)\to [0,\infty]$, namely the measure $$\mu_I(A) = \begin{cases}\infty & A\not\in I, \\ 0 & A\in I,\end{cases}$$ so that the maps $\mu \mapsto \{A\subseteq X \mid \mu(A) = 0\}$ and $I \mapsto \mu_I$ form a bijective correspondence between the $\{0,\infty\}$-valued measures on $X$ and the $\sigma$-ideals $I\subseteq \mathcal P(X)$. Give this a go if you haven't thought about it before!

Robert Trosten
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  • Nice observation! – Brian Tung Mar 08 '25 at 01:29
  • Thanks! In a past life (i.e. a past account before I lost the email) I've thought about this sort of thing for a while: https://math.stackexchange.com/questions/4679347/a-borel-measure-on-mathbbr-whose-nullsets-are-exactly-the-countable-sets – Robert Trosten Mar 08 '25 at 01:36