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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?
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    $\begingroup$ Let $A_n = [n,\infty)$. Then $\cap_n A_n = \emptyset$. $\endgroup$ – copper.hat Oct 31 '17 at 3:29
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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 $\{x\}$.

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  • $\begingroup$ 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

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