My most basic problem is that there doesn't seem to be any place containing basic theorems on transfinite sums and transfinitely-additive measures, so a reference to such a text would be very appreciated. Anyway,

My definitions:

Given a set of indices $\mathcal{I}$ and a set of positive real numbers $\{r_i\}_{i\in\mathcal{I}},$ their (possibly transfinite) sum is defined as:

$$\sum_{i\in\mathcal{I}}r_i = \sup\{\sum_{i\in I}r_i\mid I \text{ is a finite subset of }\mathcal{I}\}.$$

And, a $\lambda$-additive measure $\mu$ is a measure such that: If $\mathscr{A}$ is a family of $<\lambda$ disjoint sets in dom$(\mu)$, then $\mu(\bigcup\mathscr{A})=\sum_{A\in\mathscr{A}}\mu(A).$

I can see how to prove that given any measure a family of disjoint sets of positive measure is necessarily countable. Given this, it seems to me that all trasnfinite sums are actually countable sums and so $\sigma$-additivy would imply $\lambda$-additivy for every $\lambda$, which would make the definition superfluos, but trying to construct a hypothetical proof of this I came up with what seems to me an equivalent definition of $\lambda$-additivity which I'm not sure it is correct:

$\mu\text{ is } \lambda\text{-additive if it is }\sigma\text{-additive and}\text{ if given }\mathscr{A}\text{ a family of }<\lambda\text{ disjoint sets in dom}(\mu)$ then

$$\mu(\bigcup\mathscr{A})=\mu(\bigcup \big(\mathscr{A}\smallsetminus\{A\in\mathscr{A}\mid \mu(A)=0\})\big).$$

This definition is more intuitive to me but I keep making mistakes and I don't longer trust my verification that it is in fact equivalent. So the correctness of this definition is my first question.

Also, I'm trying to show the following for a $\mathfrak{c}$-additive measure $\mu$ on $2^\lambda$ (where $\mu(2^\lambda)=1$) for some cardinal $\lambda$: Let $(A_i)$ be a family of less than $\mathfrak{c}$ measurable sets (the $A_i$ are not necessarily disjoint) such that $\bigcup (A_i)=\ 2^\lambda$, then there exists a finite set of indices $F$ such that:

$$\mu(\bigcup_{i\in F}A_i)>\frac{7}{8}.$$

To show this, in the paper I'm reading, to prove this statement, it is said that (where $|(A_i)|=\kappa<\mathfrak{c}$):

"By $\mathfrak{c}$-additivity the transfinite sequence $(\mu(\bigcup_{i<\rho} A_i))_{\rho<\kappa}$ converges monotonically to 1."

I have not been able to apply $\mathfrak{c}$-additivity in any way as there are no disoint sets in sight and I can't see how to use $\mathfrak{c}$-additivity to show the transfinite (and possibly false) versions of measure theoretic results which would be very useful such as:

$$\mu(\bigcup A_i)=\sup_{\text{finite }F}\{\mu(\bigcup_{i\in F}A_i)\}$$ which is true for countable collections, and hence for collections of $<\mathfrak{c}$ disjoint sets, but I can't prove it is true for arbitrary $(A_i)_{i<\rho}$ and $\rho<\mathfrak{c}$.


Let $\mu:\mathcal{P}(X) \to [0, 1]$ be a countably additive measure with $\mu(X) = 1$. Let $\kappa$ be an uncountable cardinal. The following are equivalent (and we say that $\mu$ is $\kappa$-additive if they hold). Clause (2) is first definition you gave and Clause (1) is the second definition proposed by you. Clause (3) is enough to justify "By $\mathfrak{c}$-additivity ... converges monotonically to 1."

(1) For every $\theta < \kappa$ and $\{A_i : i < \theta\} \subseteq \mathcal{P}(X)$, if $\mu(A_i) = 0$ for every $i < \theta$, then $\mu(\bigcup_{i < \theta}A_i) = 0$.

(2) For every $\theta < \kappa$ and $\{A_i : i < \theta\} \subseteq \mathcal{P}(X)$, if $A_i$'s are pairwise disjoint, then $\mu(\bigcup_{i < \theta}A_i) = \sum_{i < \theta} \mu(A_i)$.

(3) For every $\theta < \kappa$ and $\{A_i : i < \theta\} \subseteq \mathcal{P}(X)$, if $A_i$'s are increasing and $A = \bigcup_{i < \theta} A_i$, then $\mu(A) = \sup \{\mu(A_i): i < \theta\}$.

(1) implies (2): Assume (1) and let $\theta < \kappa$ and suppose $\{A_i : i < \theta\} \subseteq \mathcal{P}(X)$ is a disjoint family. Since $\mu(X) = 1$, the set $W = \{i < \theta: \mu(A_i) > 0\}$ is countable. Using (1), $\mu(\bigcup_{i \in \theta \setminus W} A_i) = 0$. Hence $\mu(\bigcup_{i < \theta}A_i) = \mu(\bigcup_{i \in W}A_i) + \mu(\bigcup_{i \in \theta \setminus W} A_i) = \sum_{i \in W} \mu(A_i) + 0 = \sum_{i < \theta} \mu(A_i)$.

(2) implies (3): Assume (2) and suppose $\theta < \kappa$ and $\{A_i : i < \theta\} \subseteq \mathcal{P}(X)$ where $A_i$'s are increasing and $A = \bigcup_{i < \theta} A_i$. Define $B_i = A_i \setminus \bigcup_{j < i} A_j$ for $i < \theta$. Then $B_i$'s are pairwise disjoint. By (2), for each $i < \theta$, $\mu(A_i) = \mu(\bigcup_{j < i} B_j)$. Taking supremum over $i < \theta$, we get $\sup\{\mu(A_i) : i < \theta\} = \sup \{\mu(\bigcup_{j < i} B_j): i < \theta\} = \sum_{i < \theta} \mu(B_i) = \mu(\bigcup_{i < \theta} B_i) = \mu(A)$.

(3) implies (1): Assume (3) and suppose $\theta < \kappa$, $\{A_i : i < \theta\} \subseteq \mathcal{P}(X)$ and $\mu(A_i) = 0$ for every $i < \theta$. Put $B_i = \bigcup_{j < i} A_j$. - So $B_i$'s are increasing with $i < \theta$. It suffices to show that $\mu(B_i) = 0$ for every $i \leq \theta$. Suppose this fails and let $i_{\star} \leq \theta$ be the least $i$ for which $\mu(B_i) > 0$. But this contradicts (3) for the family $\{B_i : i \leq i_{\star}\}$.

Remark: The use of countable additivity of $\mu$ in the proof of (1) implies (2) is necessary. It is possible that there is a finitely additive extension $m:\mathcal{P}([0, 1]) \to [0, 1]$ of Lebesgue measure whose null ideal is $\mathfrak{c}$-additive but there is no countably additive total extension of Lebesgue measure.

  • $\begingroup$ When you say (for example in Clause 1) that $\mu(\bigcup_{i<\theta} A_i)=0$ is this \textit{whenever} $\bigcup_{i<\theta} A_i$ is measurable, or it part of the definition of $\kappa$-additive that $\bigcup_{i<\theta} A_i$ is always measurable? $\endgroup$ – JKEG Apr 30 '18 at 17:11

Distinguish between two topics:

(1). $r:I\to \{0\}\cup \Bbb R^+$ is a function and we write $r(i)=r_i.$ (In this case, if $i$ and $i'$ are distinct members of $I$ then $\sum_{j\in \{i,i'\}}r_j=r_i+r_{i'}$ regardless of whether or not $r_i=r_{i'}.$)

(2). $S\subset \{0\}\cup \Bbb R^+$ and define $\sum_{r\in S}r=\sup\{\sum_{r\in T}r:S\supset T \land T\text { is finite}\}.$

Topic (1). For $q\in \Bbb Q^+$ let $I_q=\{i\in I: r_i\geq q\}.$

$\Bbb Q^+$ is countable, so if $I_q$ is finite for every $q\in \Bbb Q^+$ then $\{i\in I:r_i\ne 0\} =\cup_{q\in \Bbb Q^+}I_q$ is countable.

If $I_q$ is infinite for some $q\in \Bbb Q^+$ then for any $x\in \Bbb R$ take some $n\in \Bbb N$ such that $nq>x,$ and take $J\subset I_q$ where $J$ has $n$ members. Then $\sum_{i\in I}r_i\geq \sum_{i\in J}r_i\geq nq>x.$

Topic (2). For $q\in \Bbb Q^+$ let $S_q=\{r\in S:r\geq q\}.$

If $S_q$ is finite for every $q\in \Bbb Q^+$ then $S\setminus \{0\}=\cup_{q\in \Bbb Q^+}S_q$ is countable, so $S$ is countable.

If $S_q$ is infinite for some $q\in \Bbb Q^+$ then for any $x\in \Bbb R$ take $n\in \Bbb N$ such that $nq>x,$ and take $T\subset S_q$ such that $T$ has $n$ members. Then $\sum_{r\in S} r\geq \sum_{r\in T}r\geq nq>x.$


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