Let $(\Omega, \mathcal{A}, \mu)$ be a measure space. Show that the implication $$A_n\in\mathcal{A} \;\;\;\text{and}\;\;\; A_n\downarrow A \implies \mu(A_n) \downarrow \mu(A)$$ need not be true when $\mu(A_1) = \infty$

This is the counter example I came up with. Consider the Borel measure space and the measure $\mu(A):= \begin{cases}0 & A=\emptyset \\ \#\{x\}\subset A,x\in\mathbb{Q} & otherwise \\ \end{cases} $ that is to say, the count of the number of rational singletons in $A$.

It is a measure on $\mathcal{B}$ because for any $A_n\in\mathcal{B}$, countable and pairwise disjoint, if $A_n$ contains an interval, then

$$\infty = \mu\left(\bigcup_n A_n\right) = \sum_n A_n = \infty$$

and if $A_n$ are unions of singletons then the equality above will also match.

Consider $A = \{x\}$, $x\in\mathbb{Q}$ and $A_n = [x,x+\frac{1}{n})$. Both $A$ and $A_n$ are in $\mathcal{B}$. Then $\mu(A)=1\neq\infty=\mu(A_n)$

  1. Is this correct?
  2. I feel like my counterexample is needlessly complicated. Is there a simpler counterexample?
  • 3
    $\begingroup$ Let $A_n = [n,\infty)$. Then $\cap_n A_n = \emptyset$. $\endgroup$ – copper.hat Oct 31 '17 at 3:29

The comment above has the most common counterexample of the phenomena you are talking about. To give a similar example, instead of counting rationals, just consider the measure of a set as the number of elements in that set if it is finite, and $\infty$ if the set is not finite. Then, you can consider the set you have just done i.e. for any $x \in \mathbb R$, $A_n = [x,x + \frac 1n)$ have infinite measure for all $n$, yet they decrease to a set of measure one, namely $\{x\}$.

| cite | improve this answer | |
  • $\begingroup$ Thanks. I misremember the definition of the counting measure. Was unsure if it was defined for intervals. $\endgroup$ – berrygreen Oct 31 '17 at 4:00
  • $\begingroup$ You are welcome! $\endgroup$ – Teresa Lisbon Oct 31 '17 at 4:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.