Show that some $A\in M$ attains the infimum. The following is an exercise from Bruckner's Real Analysis:

Let $(X, M, m)$ be a measure space with $m(X) = 1$, and suppose that $m(E) > 0$ for each nonempty $E \in M$. For each $x \in X$, let $ς(x) = \inf{\{m(E) : E \in M,x \in E}\}$ :
a. Show that there is a set $A_x \in M$ such that $x \in A_x$ and $m(A_x)= ς(x)$.
b. Prove that the sets ${\{A_x}\}$ are either disjoint or identical.

My attempt:
a. Let $A_x=\cap_{i=1}^{\infty} E_i$ in which $E_i$ is measurable and contains $x$. It is again measurable so equals the infimum. The problem with this is that how to prove ${\{E_i}\}$ is countable?
b. Define $x~y$ iff $x,y \in E_i$ for some $i$. Then it's an equivalence relation and partitions $X$ and if some collection $E_i$ is disjoint with another collection $F_i$ so is their intersection.
My questions:
1- Are my arguments correct? and why?
2- Why $m(E) > 0$ is necessary in the hypothesis of the theorem?
3- Why $m(X) =1$ is necessary in the hypothesis of the theorem?
 A: The sets which contain $x$ may be uncountably many, so your statement (a) is not correct at the very beginning.
Anyway, you do not have to select all such sets. Rather, for each $n=1,2,...$, pick an $E_{n}$ such that $\varsigma(x)\leq m(E_{n})\leq\varsigma(x)+1/n$. Take $A_{x}=\bigcap_{n}E_{n}$, then $m(A_{x})=\lim_{n}m(E_{n})$ as the measure space is of finite measure. Therefore, $m(A_{x})=\varsigma(x)$ as claimed.
EDIT:
Consider any $A_{x},A_{y}$, we have two cases.
Case I.
If $y\in X-A_{x}$, then $y\in A_{y}\cap(X-A_{x})$. However, $m(A_{y})$ being defined as the infimum, we have $m(A_{y})\leq m(A_{y}\cap(X-A_{x}))$, then $m(A_{y}-(A_{y}\cap(X-A_{x})))=0$. The assumption $m(B)>0$ for nonempty $B$ implies that $A_{y}-(A_{y}\cap(X-A_{x}))=\emptyset$ and hence $A_{y}\subseteq X-A_{x}$. In this case, we have $A_{x}\cap A_{y}=\emptyset$.
Case II.
If $y\notin(X-A_{x})$, in other words, $y\in A_{x}$, then $y\in A_{x}\cap A_{y}$. And again by the infimum, $m(A_{y})\leq m(A_{x}\cap A_{y})$. The same trick implies $A_{y}\subseteq A_{x}\cap A_{y}\subseteq A_{x}$. In this case, we have $A_{y}\subseteq A_{x}$.
On the other hand, if we have $x\in X-A_{y}$, by the reasoning given in Case I, we see that $A_{x}\cap A_{y}=\emptyset$, but this contradicts the earlier fact that $y\in A_{x}\cap A_{y}$. Hence, $x\notin X-A_{y}$, and hence $x\in A_{y}$. And again by the previous paragraph, we deduce that $A_{x}\subseteq A_{y}$.
In this Case II, we obtain that both $A_{y}\subseteq A_{x}$ and $A_{x}\subseteq A_{y}$, so $A_{x}=A_{y}$.
