I've searched and haven't seen this problem in my searches. Admittedly I may have missed it and I apologize if that's the case.

The problem statement is the following:

Let $(X, \mathcal{M}, \mu)$ be a measure space of infinite measure, with the following property: for every $A \in \mathcal{M}$ with $\mu(A) = \infty$ there exists $B\subset A$ with $0 < \mu(B) < \infty$. Show that there exist sets $E_i\in M(i\geq 1)$ such that $\mu(E i ) < \infty$ (for all $i\geq 1$) and $\mu(\cup_{i\geq1}E_i ) = \infty$.

Given the property that there exists $B \subset A \text{ with } 0<\mu(B)<\infty$ it's easy to see that you can pull an infinite number of subsets $B$ from $A$ since $\mu(A\setminus B) = \infty$. However, given that $\mu(B)$ is simply a finite value greater than zero I'm having trouble showing that $\mu(\cup_{i\geq1}E_i)=\infty$ because $\mu(E_i)$ could be such a small value for every $E_i$ that the measure of the union is still finite.

I feel like it may require a proof by contradiction, but I'm unsure how to proceed with it.


In fact, we can choose $\mu(E_i)$ to be as large as we desire. More concretely, given the measure space with the property you mention, if $A\in\mathcal M$ and $\mu(A)=\infty$, then for any $C>0$ there exists $B\subset A$ with $C<\mu(B)<\infty$. To see this, let $M=\sup\{\mu(B):B\subset A\text{ and }\mu(B)<\infty\}$. If $M=\infty$ the result follows. Otherwise if $M$ is finite, then for each $n$ we can choose $B_n\subset A$ such that $M-\frac{1}{n}<\mu(B_n)\leq M$. Put $B=\cup_nB_n$, so that $\mu(B)=M$. But then $\mu(A\setminus B)=\infty$, so there is some $B'\subset A\setminus B$ with $0<\mu(B')<\infty$, and $M<\mu(B\cup B')<\infty$, contradicting the definition of $M$.

By the above, there exists $E_1\in\mathcal M$ such that $1<\mu(E_1)<\infty$. Then define $E_i$ inductively: There exists $E_{i+1}\subset X\setminus(E_1\cup\cdots\cup E_i)$ with $1<\mu(E_{i+1})<\infty$. Thus $\mu(E_i)$ is finite for all $i$, and since the $E_i$ are disjoint we have $\mu(\cup_iE_i)=\sum_i\mu(E_i)=\infty$.

  • $\begingroup$ Thanks for catching that $\endgroup$ – Aweygan Sep 20 '18 at 16:31
  • $\begingroup$ I don't see how you show that there is $E_1$ of finite measure larger than 1. This seems to be the whole issue. $\endgroup$ – Andrés E. Caicedo Sep 20 '18 at 16:40
  • 1
    $\begingroup$ From the result I state in the first paragraph, with $C=1$. Should I include a proof? $\endgroup$ – Aweygan Sep 20 '18 at 16:44
  • $\begingroup$ I think the current version is quite clear, thank you. $\endgroup$ – Andrés E. Caicedo Sep 20 '18 at 16:51

Note first that if $C$ is uncountable and for each $i\in C$ we have a real number $r_i>0$, then $\sum_i r_i=+\infty$. The point is that for some $n>0$, the family $\{i\in C: r_i>1/n\}$ is uncountable. Of course, in this case there is a countable subfamily whose sum is infinite as well.

Now, let $A$ be measurable of infinite measure. If there is an uncountable disjoint family of measurable subsets of $A$ of strictly positive finite measure, we are done by the observation in the last paragraph. It follows that we may assume that any such family is countable. Use Zorn's lemma to argue that there is a maximal such family $\mathcal F$. Since it is countable, $A\smallsetminus \bigcup_{B\in \mathcal F}B$ is measurable, and must have measure 0. It follows that $\mu(\bigcup_{B\in\mathcal F} B)=+\infty$, and we are done. Note we did a bit more than asked for in the question, as we arranged to find the family of sets inside any given set of infinite measure.

(Rather than Zorn's lemma, I would prefer to build the family by transfinite induction. If you are familiar with this technique, I think it is preferable, as you do not have to worry about arguing that the conditions of Zorn's lemma are met, which is why I had the separate into two cases depending on whether there was an uncountable disjoint family. The issue is that we can only ensure measurability of countable unions of measurable sets, so some care is required.)


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