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.
Best Answer
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 $\mu(\cup_iE_i)=\sum_i\mu(E_i)=\infty$.