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I have been given the following definitions:

A measure $\mu$ is a weighted counting measure on $(X,\mathscr{M})$ if there exists a function $f:X\rightarrow[0,\infty]$ such that $$\forall E\in\mathscr{M}, \mu(E)=\sum_{x\in E} f(x)$$

A non-null set $A\in\mathscr{M}$ is $\mu$-atomic if for each $B\in\mathscr{M}$ with $B\subseteq A$, either $\mu(B)=0$ or $\mu(B)=\mu(A)$. If every non-null $E\in\mathscr{M}$ contains an atom then we say $\mu$ is purely atomic

My professor proposed the following problem as just something fun to think about, not as a real homework question:

For any measurable space $(X,\mathscr{M})$, are all weighted counting measures purely atomic?

After some thought, I believe that this question reduces to the case when $\mu$ is a finite measure, but I am struggling to make much further progress. Can anyone help me out?

BoredAtWork100
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  • I'm confused. A set containing one element, say $A = {x}$, is always an $\mu$-atomic. With this definition, it seems that almost any measure will satisfy the "purely atomic" definition, or perhaps none if you allow empty sets. – Surge Nov 17 '23 at 16:45
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    @Surge Atoms are usually required to have nonzero measure – FShrike Nov 17 '23 at 16:47
  • To check the definition of a purely atomic measure, we should exclude the null sets. This problem might become easier for you if you consider $\mu=\sum_{x\in X}f(x)\cdot\delta_x$ – FShrike Nov 17 '23 at 16:51
  • You should probably demand that $E$ contains an atom only if it is not a null set. Otherwise, the Dirac measure on the Borel sigma algebra would not be purely atomic. – Keen-ameteur Nov 17 '23 at 16:54
  • @FShrike My apologies. You are correct in stating the atoms should be required to be non-null and that we should exclude null sets when checking the definition of purely atomic. I will update the definition accordingly – BoredAtWork100 Nov 17 '23 at 17:08

1 Answers1

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Let $(X,\mathcal{M},\mu)$ be measure space and $f:X\to[0,\infty]$ a function such that $\displaystyle\mu(A)=\sum_{x\in A}f(x)$ for any $A\in\mathcal{M}$.

Suppose $\mu(A)=\infty$.

If every $B\in\mathcal{M}$ with $B\subseteq A$ has either measure $0$ or infinite measure, then, by definition, $A$ is atomic.

Otherwise take $B\in\mathcal{M}$ with $B\subseteq A$ and $0<\mu(B)<\infty$. It will follow from the finite measure case that $B$ contains an atom. Since $B\subseteq A$ , $A$ contains an atom.

Now suppose $0<\mu(A)<\infty$.

For $M\in\mathcal{M}$ consider the notation $M^+:=\{x\in M:f(x)>0\}$. Notice that if $B\subseteq A$ and $\mu(B)<\mu(A)$, then $B^+\subsetneq A^+$. (otherwise we would have the same finite sums and therefore the same supremum of finite sums.)

Since $\mu(A)<\infty$, the set $A^+$ is countable$^{(*)}$. Let $\{x_0,x_1,x_2,...\}$ be an enumeration of that set.

For each $n\in\mathbb{N}$ let $S_n\in\mathcal{M}$ be a set that separates $x_0$ and $x_n$, more precisely, a set such that $x_n\in S_n$ and $x_0\notin S_n$. If there is no such set define $S_n=\emptyset$.

Define $\displaystyle S=\bigcup_{n\in\mathbb{N}}S_n$. If $x_n\notin S$ then no measurable set separates $x_0$ and $x_n$, otherwise we would have $x_n\in S_n\subseteq S$, and clearly $x_0\notin S$.

The set $A\setminus S$ is atomic:

Since $x_0\in A\setminus S$, we have $\mu(A\setminus S)\geq f(x_0)>0$, so $A\setminus S$ has positive measure.

Now suppose, for contradiction, that there is a set $B\in\mathcal{M}$ such that $B\subseteq A\setminus S$ and $0<\mu(B)<\mu(A\setminus S)$.

Since $\emptyset\subseteq B\subseteq A\setminus S$ and $0<\mu(B)<\mu(A\setminus S)$ we have $\emptyset\subsetneq B^+\subsetneq (A\setminus S)^+=(A^+\setminus S)$. But then $B^+$ separates $x_0$ from the elements of $(A^+\setminus S)\setminus B^+$, which is a contradiction.

$(*)$ If $A^+$ is uncountable, then $\mu(A)=\infty$:

Define $E_k:=\left\{x\in A:f(x)>\frac{1}{k}\right\}$. We can write $\displaystyle A^+=\bigcup_{k\in\mathbb{N}}E_k$. Since $A^+$ is uncountable, some $E_k$ is uncountable, in particular, some $E_k$ is infinite.

Since there are infinitely many $x\in A$ with $f(x)>\dfrac{1}{k}$, it follows that $\mu(A)=\infty$.

If we instead considered a set to be atomic if $$ B\subseteq A~,~B\in\mathcal{M} \implies \mu(B)=0\text{ or }\mu(A\setminus B)=0 $$

then the $\mu(A)=\infty$ case no longer works. But adding the assumption that $\mu$ is $\sigma$-finite we can still take a subset of $A$ with finite measure.

And if $X$ is countable, then $A^+$ is always countable regardless of measure, and the second case applies directly.

With this definition the result is false in general. A counterexample is the $\sigma$-algebra given in this answer by taking $f\equiv 1$.

VPPVPVVP
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