6

My task is to prove that if an atomic measure space is $\sigma$-finite, then the set of atoms must be countable.

This is my given definition of an atomic measure space:

Assume $(X,\mathcal{M},\mu)$ is a measure space with all single points being measurable. An atom is a point $x$ with $\mu(\{x\}) > 0$. Letting $\mathcal{A}$ be the set of atoms, $(X,\mathcal{M},\mu)$ is called atomic if $\mathcal{A}\in\mathcal{M}$ and $\mu(\mathcal{A^c}) = 0$.

I didn't know how to prove this at first, so I looked it up on stack exchange and found this answer: (I do not have enough reputation to comment on the original post)

Here's how to prove your claim, with the appropriate assumption. Let $S\subset X$ be the set of atoms for some measure $\mu$ on $X$. Let $\{U_i\}$ be a countable measurable partition of $X$. Then if $S$ is uncountable, some $U_i$ contains an uncountable subset $S'$ of $S$, and $\mu(U_i)\geq \sum_{x\in S'}\mu(x)=\infty$ since any uncountable sum of positive numbers diverges. Thus $\mu$ is not $\sigma$-finite.

My question is why do we have that $\mu(U_i) \geq \sum_{x\in S'} \mu(x)$ ? I am assuming that this inequality comes from subadditivity of $\mu$ but as I have understood it subadditivity is defined for countable unions, not for uncountable unions so I am confused as to how we arrive at an uncountable sum in this step.

Axel Andersson
  • 71
  • 2
    Have a look at this https://math.stackexchange.com/questions/20661/the-sum-of-an-uncountable-number-of-positive-numbers for some relevant discussion of what uncountable sums of non-negative numbers mean. Subadditivity and the discussion there shows that $U_i$ must contain a countable subset of $S'$ that has infinite measure, implying that $U_i$ has infinite measure. I agree that the proof you have quoted could usefully have explained this. – Rob Arthan Sep 28 '21 at 22:32

3 Answers3

3

That does appear to be a bit sloppy. But you can mend that by simply specifying the sum over any countable selection and taking the supremum of that. You can easily show: If you have an uncountable family of positive values, then for any countable finite sum there exists an even larger one, so this sup must be $\infty$ (for else take a sequence of such countable selections so that the sum converges to the sup. Then the union of all these selections is also a countable selection, and at least as large as the limit).

Lazy
  • 4,519
  • I see, so is it valid in this context that "The set of atoms must be countable" is equivalent to "if there are uncountably many singletons in $\mathcal{M}$ with positive measure (atoms) then there is no way to cover $X$ with $\mu$-finite subsets using a countable union"..? – Axel Andersson Sep 30 '21 at 19:08
  • 1
    Not equivalent, no. You can also have problems with a countable amount of atoms (e.g. when some atom has measure $\infty$ or there is a limit point of atoms where the sum of the measures diverges.

    But the idea is: If there are uncountably many atoms there has to be a set in any countable partition that contains uncountably many of these. But any uncountable set of singletons with positive mass has countable subsets with arbitrarily large measure. So the set cannot have finite measure.

     – Lazy Sep 30 '21 at 19:20
2

You may argue as follows. Since $U_i\supset S'$ for some $i\in\mathbb{N}$, $$ \mu(U_i)\ge \mu(S'')\label{1}\tag{1} $$ for any $S''\subseteq S'$. Let $\mathcal{S'}$ be the collection of all sequences of points in $S'$. Then, $$ \mu(U_i)\ge \sup_{s\in\mathcal{S'}}\sum_{x\in s}\mu(\{x\})\equiv\sum_{x\in S'}\mu(\{x\})=\infty, $$ where the inequality follows from \eqref{1} and countable additivity, and the last equality holds because there exists $\epsilon>0$ s.t. $\mu(\{x\})\ge \epsilon$ for uncountably many $x$'s in $S'$.

1

Here I will take "countable" to mean finite or countably infinite.

For $n\in\{1,2,3,\ldots\},$ if the set $A_n = \{ x : \mu(\{x\}) \ge 1/n \}$ were not countable then the space would not be $\sigma$-finite.

Since $(0,+\infty)= \bigcup_{n=1}^\infty \left(\frac 1 n, +\infty\right),$ the set $\bigcup_{n=1}^\infty A_n$ is the set of all atoms. This is a countable union of countable sets.

Michael Hardy
  • 1