Uncountably additive measures

$\newcommand{\F}{\mathscr F}$$\newcommand{\a}{\alpha}$$\newcommand{\N}{\mathbb N}$$\newcommand{\w}{\omega}$$\newcommand{\E}{\mathcal E}$In this post I’m going to explore uncountable additivity of $\sigma$-finite measures a bit.

Let $\mathcal M := (\Omega, \F)$ be a measurable space and let $\mu : \F \to [0,\infty]$ be a measure so it gives the sizes of the subsets of $\Omega$ contained in the $\sigma$-algebra $\F$. Part of the definition of a measure is that it is countably additive, i.e. if $\{E_1, E_2,\dots\}$ is a pairwise disjoint collection of sets in $\F$ then
$$
\mu\left(\bigcup_{n=1}^\infty E_n\right) = \sum_{n=1}^\infty \mu(E_n).
$$
If $\Omega$ can be partitioned into a disjoint countable collection $\{E_1,E_2,\dots\}$ such that for each $n\in\N$ $\mu(E_n) < \infty$ then $\mu$ is $\sigma$-finite. I will let $\E$ represent an arbitrary such collection, and WLOG I’ll further take $\mu(E) > 0$ for each $E \in \E$.

My goal here is to see what happens if I allow $\mu$ to be uncountably additive, which I’ll define as follows: if $\{E_\a : \a \in A\}$ is an uncountable pairwise disjoint collection then
$$
\mu\left(\bigcup_{\a \in A} E_\a\right) = \sum_{\a \in A} \mu(E_\a).
$$
Throughout this I’ll be assuming that $\Omega$ is uncountable otherwise the difference between countable and uncountable additivity never matters. I’ll also be assuming that each singleton $\{\w\}$ for $\w \in \Omega$ is measurable, i.e. $\{\{\w\} : \w \in \Omega\}\subset \F$.

The first order of business is to work out what $\sum_{\a \in A} \mu(E_\a)$ means as this can be the sum of uncountably many things.

I’ll follow what’s done in Tao’s An Introduction to Measure Theory.

Uncountable sums

Let $\{x_\alpha : \alpha \in A\} \subseteq [0, +\infty]$ be an arbitrary indexed collection. I’ll define the sum $\sum_{\alpha \in A} x_\alpha$, which could have uncountably many terms, by
$$
\sum_{\alpha \in A} x_\alpha := \sup_{F \subset A, |F| < \infty} \sum_{\alpha \in F} x_\alpha.
$$
If $A = \emptyset$ then I’ll take $\sum_{\a \in A}x_\a = 0$. This definition does imply that the order of summation doesn’t matter, but since I’m only summing non-negative things that won’t be a problem for me. Another point is that because the extended reals are complete, this $\sup$ always exists (although it may be infinite) which means that every such sum converges.

This has a consequence for the measurability of subsets of $\Omega$: if I let $E \subseteq \Omega$ then by uncountable additivity I have
$$
\mu(E) = \sum_{\w \in E} \mu(\{\w\}).
$$
That sum is guaranteed to converge although it may be infinite, but regardless I am able to consistently assign a measure to every element of $2^\Omega$ so I’ll just take $\F = 2^\Omega$ and I won’t worry about measurability.

Now I’ll prove two results about this new notion of summation.

Result 1: if $A$ is countable this agrees with the usual notion of summation.

Pf: WLOG I can take $A$ to be $\N$ so I am summing $\{x_n : n \in \N\}$. Since the elements being summed are all nonnegative I don’t need to worry about the order given by this particular enumeration of $A$.

I want to show
$$
\sup_{F\subset\N, |F| < \infty} \sum_{k \in F}x_k = \lim_{n\to\infty} \sum_{k =1}^n x_k.
$$
Letting $J_n = \{1,\dots,n\}$ this is equivalent to showing
$$
\sup_{F\subset\N, |F| < \infty} \sum_{k \in F}x_k = \lim_{n\to\infty} \sum_{k \in J_n} x_k.
$$
For the right-hand side, because each $x_k \geq 0$ I have
$$
\sum_{k \in J_{n+1}}x_k = x_{n+1} + \sum_{k \in J_n} x_k \geq \sum_{k\in J_n} x_k
$$
so the partial sums $\sum_{k\in J_n} x_k$ are monotonic and bounded above by $+\infty$ which means that they converge, but also that the limit is equal to the supremum of the set. Thus
$$
\lim_{n\to\infty}\sum_{k\in J_n} x_k = \sup_{n\in\N}\sum_{k\in J_n} x_k.
$$
Since every $J_n$ is a finite subset of $\N$ this means that every $\sum_{k\in J_n} x_k$ appears as some $\sum_{k \in F} x_k$, so
$$
\sup_{F\subset\N, |F| < \infty} \sum_{k \in F}x_k \geq \lim_{n\to\infty} \sum_{k \in J_n} x_k.
$$
For the other direction, let $F \subset \N$ be an arbitrary finite index set. Then there is some $n\in \N$ such that $F \subseteq J_n$ which means that every element of $\left\{\sum_{k \in F} x_k : F\subset \N \text{ is finite}\right\}$ is bounded above by some element of $\left\{\sum_{k \in J_n} x_k : n \in \N\right\}$ so
$$
\sum_{k \in \N} x_k = \sup\left\{\sum_{k \in F} x_k : F\subset \N \text{ is finite}\right\} \\
\leq \sup \left\{\sum_{k \in J_n} x_k : n \in \N\right\} = \sum_{k=1}^\infty x_k.
$$

Thus the two notions of summation agree in the case with the index set $A$ being countable.

$\square$

This next result will be very important.

Result 2: $\sum_{\a \in A} x_\a < \infty$ only if all but countably many $x_\a$ are zero (even if $A$ is uncountable). This is still assuming that $x_\a \geq 0$ for all $\a \in A$. Pf: For each $n \in \N$ let $A_n = \{\a : x_\a > \frac 1n\}$. If $\sum_{\a \in A} x_\a < \infty$ then it must also be that $\sum_{\a \in A_n} x_\a < \infty$ for each $n \in \N$. Since $\infty > \sum_{\a \in A_n} x_\a > \sum_{\a \in A_n} \frac 1n$ it must be that $A_n$ is finite for each $n$. This means that
$$
\{\a : x_\a > 0\} = \bigcup_{n \in \N} A_n
$$
which is a countable union of finite sets and is therefore countable.

$\square$

Now I can use this notion of summation to reason about uncountable additivity and since it agrees with the usual infinite summation in the countable case I don’t need to worry about having somehow messed up what $\sigma$-additivity means.

I have my partition $\E$ of $\Omega$ into at most countably many pieces each with positive finite measure. $\Omega$ is uncountable so at least one $E_n$ is uncountable, but by uncountable additivity and the singletons $\{\w\}$ being measurable, I have
$$
0 < \mu(E_n) = \sum_{\w \in E_n} \mu(\{\w\}) < \infty
$$
for each $n\in\N$. By Result 2 this means that no matter the cardinality of $E_n$ at most countably many of the $\{\w\}$ in it have positive measure, and that at least one of them has positive measure. For each $E_n$ I’ll denote these by $S_n := \{\w_{n1},\w_{n2},\dots\}$. Also since $S_n, E_n \in \F$ I know $\mu(E_n \backslash S_n) = 0$ so $\mu$ is entirely supported by the countable set $S_n$ in $E_n$.

Then because a countable union of countable sets is countable, this means that
$$
S := \bigcup_{n \in \N} S_n
$$
is an at most countable set of elements with positive measure and $\Omega \backslash S$ has measure zero. Also for $S$ the order of the collection $S$ is irrelevant so the particular partition $\E$ doesn’t matter any more.

Let $F\subseteq \Omega$. Then
$$
\mu(F) = \mu(F \cap S) + \mu(F \cap S^c) = \mu(F\cap S)
$$
so any set’s measure is determined by which elements of $S$ it contains.

At this point I’ve shown that uncountable additivity (along with the singletons all being measurable) means that any uncountably additive measure is necessarily entirely supported on a countable collection of singleton atoms.

This rules out the Lebesgue measure as being uncountably additive, which is also trivially seen by
$$
1 = \lambda([0,1]) \neq \sum_{x \in [0,1]} \lambda(\{x\}) = 0.
$$

Here’s an example of an uncountably additive probability measure.

Take $\Omega = \mathbb R$ and $\F = 2^{\mathbb R}$. Let $\{q_1,\dots\}$ be an enumeration of $\mathbb Q$ and define a measure $\nu$ by
$$
\nu(E) = \sum_{n \,:\, q_n \in A} 2^{-n}
$$
so $\nu$ gives an exponentially decaying sequence of point masses on the rational numbers. This makes $\nu$ exactly of the form mentioned in the previous section (i.e. supported on a countable set with $\mathbb Q = S$).

First I’ll show this is indeed a measure. Clearly for all $E\in \F$ I have $\nu(E) \geq 0$ and $\nu(\emptyset) = 0$. Now let $E_1, E_2,\dots$ be a disjoint collection in $\F$. I have
$$
\sum_{k \in \N} \nu(E_k) = \sum_{k\in\N}\sum_{n \,:\,q_n \in E_k} 2^{-n}.
$$
Because the $E_k$ are disjoint this is exactly the sum of $\{2^{-n} : q_n \in \cup_k E_k\}$, and since all the terms are non-negative I can rearrange this sum to be exactly that, i.e.
$$
\sum_{k \in \N} \nu(E_k) = \sum_{k\in\N}\sum_{n \,:\,q_n \in E_k} 2^{-n} = \sum_{n\,:\, q_n \in \cup E_k} 2^{-n} = \nu(\cup_k E_k)
$$
so $\nu$ is indeed a valid measure.

Now to show that it is in fact uncountably additive: let $\{E_\a : \a \in A\}$ be an uncountable disjoint collection in $\F$ and take $E = \cup_\a E_\a$.

Let $B = \{\a \in A : E_\a \cap \mathbb Q \neq \emptyset\}$ and note that if $\a \notin B$ then $\nu(E_\a) = 0$ since it contains no rational values. Since the $E_\a$ are disjoint and $\mathbb Q$ is countable, $B$ must also be countable. This means

$$
\sum_{\a \in A} \nu(E_\a) = \sum_{\a \in B} \nu(E_\a) = \nu\left(\bigcup_{\a \in B} E_\a\right)
$$
by countable additivity. Then since $\cup_{\a \notin B} E_\a$ is disjoint from $\cup_{\a \in B} E_\a$ and has measure zero,
$$
\nu\left(\bigcup_{\a \in B} E_\a\right) = \nu\left(\bigcup_{\a \in A} E_\a\right)
$$
so $\nu$ is uncountably additive.

$\square$

So, in conclusion, if I want an uncountably additive $\sigma$-finite measure I’ll need to consider measures that are supported on a countable collection of point masses, and this makes sense because then they’re not really even uncountably additive because the uncountable part just gets shoved into a set of measure zero and countable additivity really is what’s doing the work on the supporting singletons.;