Doubt about proof Hahn decomposition 
Theorem. Let $\mu$ a signed measure on sigma-algebra $\mathcal{A}$ such that $-\infty<\mu(E)\le+\infty$ where $E\in\mathcal{A}$. Then $X=H^+\cup H^-$, $H^+\cap H^-=\emptyset$, where $H^+$ is a positive set for $\mu$ and $H^-$ is a negative set for $\mu$

My doubt concerns the beginning of the proof.
proof. Let $$\beta=\inf\{\mu(B):B\in\mathcal{A},\;B\;\text{negative}\}.$$ We consider a negative sequence of sets $\{B_n\}$ such that $\beta=\lim_{n\to\infty}\mu(B_n)$.
Question. Can we always find a sequence of this kind? If yes, why? Does it focus on the fact that subsets of negative sets are negative?
Thanks!
 A: As asdf said, by definition of "infimum".  More details:  
Given $n$, the number $\beta+\frac{1}{n}$ is not a lower bound of the set $\{\mu(B) :B\in\mathcal{A},\;B\;\text{negative}\}$.  Therefore there is a negative set $B_n \in \cal A$ with $\mu(B_n) < \beta+\frac{1}{n}$.  And of course, since $B_n$ is a negative set, $\mu(B_n) \ge \beta$.  So $|\mu(B_n) - \beta| < \frac{1}{n}$.  
No, we did not use the fact that a subset of a negative set is a negative set.
A: Yes. More generally, if $A \subseteq \mathbb{R}$ is a nonempty set with infimum $a = \inf(A)$, then there is a sequence $(x_n) \in A$ such that $(x_n)\to a$. The same result holds for a supremum.
In your case, you're looking at a (nonempty) set $A = \{\mu(B): B\in\mathcal{A}, B\text{ negative}\}$. Since $A$ has an infimum $\beta$, it follows from the previous paragraph that there is a sequence $(x_n)$ in $A$ converging to $\beta$. But since each $x_n\in A$, for each $n\in\mathbb{N}$ there exists a negative set $B_n\in\mathcal{A}$ such that $x_n = \mu(B_n)$.
A: This is a consequence of the definition of infimum - if such sequence doesn't exist then you can increase it.
