Problem Prove that if $\mu$ is a semifinite measure and $\mu(E) = \infty$, then for every $C > 0$ there exists $F \subset E$ with $C < \mu(F) < \infty$.
My answer We can define a disjoint "chain" of sets by letting $F_n$ be the finite set of nonzero measure lying inside $E - F_1 - F_2 - \cdots - F_{n - 1}$. The "weight" of the chain $F$ is defined as $\sum_{i = 1}^{\infty} \mu(F_i)$. Let $W$ be the set consisting of the weight of every possible chains with the finite weight.
Now assume that $\sup W = s < \infty$. For any $n \in \mathbb{N}$ there exists a chain $C_n$ such that the weight is heavier than $s - \frac{1}{n}$. We can join every $C_n$ using Cantor diagonalization to make a new chain $D'$, but it is not assumed to be disjoint. Let a chain $D$ defined as follows: $$\begin{gather}D_1 = D'_1 \\ D_n = D'_n - \bigcup^{\infty}_{i = 1} D'_i. \end{gather}$$
The chain $D$ includes every $C_n$, and therefore its weight is $s$. Let $E = \bigcup_{i = 1}^{\infty} D_i$. It is clearly inside the domain of $\mu$, and $\mu(E) = s$. But then there exists some $F \subset E^c$ with $0 < \mu(F) < \infty$, and the chain constructed by attaching $F$ to $D$ has a weight heavier than $s$. So we may conclude that $s = \infty$.
Now for every $C > 0$ there exists a chain with a weight heavier than $C$, and the union of all the sets of that chain is the set with a measure greater than $C$.
How does it look? Is there any errors?
Thanks in advance.