I was reading through the proof of the Hahn Decomposition Theorem on Wikipedia. The proof starts with a claim:
Suppose that $D\in\Sigma$ satisfies $\mu(D)\leq 0$. Then there is a negative set $A\subset D$ such that $\mu(A)\leq\mu(D)$.
The sketch of proof is roughly taking away disjoint and positively measured subsets $B_n$'s of $D$ to arrive at the negetive set $A:=D \backslash \bigcup_{n=0}^{\infty} B_{n}$. The subsets $B_n$'s are constructed inductively. Assume $B_0,B_1,B_2,\dots,B_{n-1}$ have been constructed and $A_n:=D\backslash\bigcup_{m=0}^{n-1}{B_m}$, then $B_n$ is chosen to be large enough in the sense of $$ \mu\left(B_{n}\right) \geq \min \left(1, \frac{t_{n}}{2}\right), $$ where $$ t_{n}:=\sup \left(\left\{\mu(B) \mid B \in \Sigma \text { and } B \subseteq A_{n}\right\}\right). $$
The question is:
Why $\mu(B_n)\geq\min\left(1,\frac{t_n}{2}\right)$ instead of simply $\mu(B_n)\geq \frac{t_n}{2}$? How does the minimum term work?