Show that there is $E \in M$ such that $\mu(E)=\alpha$ for any $0 \lt \alpha \lt \mu(X)$ Let $(X,M,\mu)$ be a finite measure space with no atoms. A set $A \in M$ is called an atom if $\mu(A) \gt 0$ and for any measurable subset $B \subset A$ ,either $\mu(B)=0$ or $\mu(A-B)=0$ .Show that there is $E \in M$ such that $\mu(E)=\alpha$ for any $0 \lt \alpha \lt \mu(X)$.
My attempt:
For any $F \in M$, there exists a $B \in F$ such that $\mu(B) \ne 0$ and $\mu(F-B) \ne0$ (Measure Space has no atoms).Since $\mu$ is finite, there exists a $k_0 \in \mathbb{N}$ such that $\mu(F) \lt k_0$. Then  since $\mu(F)=\mu(B)+\mu(F-B)$. Atleast one of $B$ or $F-B$ has measure less than  $\dfrac{\mu(F)}{2}$. Call it $B_1$. Similarly we can find $B_2 \subset B_1$ such that $\mu(B_2) \lt \dfrac{\mu(F)}{2^2}$. By induction we can find $B_n$ such that $B_n \subset B_{n-1}\subset....\subset B_2 \subset B_1$ such that $\mu(B_n)\lt \dfrac{\mu(F)}{2^n}$. This just means that given $\epsilon \gt 0$ , there is a set $S \in M$ such that $\mu(S) \lt \epsilon$. 
Now I try to define a set $$T=\{E \in M| \mu(E) \le \alpha\}$$. This set is nonempty by the above observation. I define the order relation $"\le"$ by inclusion. Then $T$ is a partially ordered set. Any linearly ordered set $L$ in $T$ has sets which are nested. The upper bound is the union of all these, which is unfortunately not in $M$, for the union can be be over any index, countable and uncountable. 
This makes me to change the set to $$T'=\{E_{\alpha} \in M|  \mu(E_{\alpha}) \le \alpha,\mu(E_{\alpha}\cap E_{\beta})=\phi, \text{for} ,\alpha, \beta \in I, \alpha \ne \beta\}$$
Then since $\mu(X)\lt \infty$, there are atmost countably many of these sets as defined in $T'$. But I am unable to find an order relation for this set. Another way could be to define an order relation such that any linearly ordered set has of the form $T'$. 
I tried another possible approach:
Define ~ on $ M $ by $ E $~ $F$ iff $\mu ((E\cap F^{c}) \cup (E^{c}\cap F^{c}))=0$. Then this is an equivalence relation on $ M $ and defining $ d ([E], [F])= \mu ((E\cap F^{c}) \cup (E^{c}\cap F^{c})$ turns $M $ into a metric space. Now the problem just boils down to showing that the space $( M$,~$) $ is connected with the above metric. 
Thanks for the help!! 
 A: I came up with this, using the nuts and bolts of sigma-algebras instead:
Lemma. With $(X,M,\mu)$ as in the problem statement, assume $A\in M$, $\mu(A)>0$.
Then there exists $A'\in M$ with $A'\subset A$ and $\frac14\mu(A)<\mu(A')<\frac34\mu(A)$.
Proof.
Let $S=\{\,B\in M\mid B\subset A, \mu(B)\ge\frac 12\mu(A)\,\}$. As $A$ is not an atom, $S\ne \emptyset$ and we can consider $s:=\inf\{\,\mu(B)\mid B\in S\,\}$. Clearly, $\frac 12\mu(A)\le s<\mu(A)$.
Assume  $s\ge\frac34\mu(A)$.
For $B,B'\in S$, we have 
$$ \mu(B\cap B')\ge\mu(B)+\mu(B')-\mu(A)\ge 2s-\mu(A)\ge\frac12\mu(A),$$
i.e., $S$ is closed under finite intersections.
For $n\in\Bbb N$, pick $B_n\in S$ with $\mu(B_n)<s+\frac1n$.
Let $C=\bigcap B_n$. As $\bigcup_{n=1}^NB_n\in S$ and $s\le \mu(\bigcup_{n=1}^NB_n)<s+\frac 1N$, we infer
$$\mu(C)=\lim_{N\to\infty}\mu\biggl(\bigcup_{n=1}^NB_n\biggr)=s.$$
As $C$ is not an atom, there is $D\in M$ with $D\subset C$ and $0<\mu(D)<\mu(C)$. From $\mu(D)<s$, we conclude $\mu(D)<\frac 12\mu(A)$, $\mu(A-D)>\frac12\mu(A)$, $\mu(A-D)\ge s$, $\mu(D)\le\mu(A)-s$, $\mu(C-D)\ge 2s-\mu(A)\ge \frac12\mu(A)$, $\mu(C-D)\ge s$, contradiction.
We conclude that $s<\frac34\mu(A)$.
Hence we can pick $A'\in S$ with $\mu(A')<\frac34\mu(A)$. $\square$
Let $\alpha$ be given with $0<\alpha<\mu(X)$.
We construct sequence $A_0,A_1,\ldots$ and $B_0,B_1,\ldots$ as follows, where we ensure $A_n\subseteq A_{n+1}\subsetneq B_{n+1}\subseteq B_n$ at each step:
Let $A_0=\emptyset$, $B_0=X$.
Given $A_n\subsetneq B_n$, use the lemma to pick $A'\subset B_n-A_n$ with $\frac 14\mu(B_n-A_n)<\mu(A')<\frac 34\mu(B_n-A_n)$.
If $\mu(A_n\cup A')<\alpha$, let $A_{n+1}=A\cup A'$, $B_{n+1}=B_n$; otherwise let $A_{n+1}=A_n$, $B_{n+1}=A_n\cup A'$.
Observe that $\mu(A_n)\le\alpha\le \mu(B_n)$ for all $n$.
We also have $\mu(B_{n+1}-A_{n+1})<\frac34\mu(B_{n}-A_{n})$.
Therefore with $A=\bigcup A_n$ and $B=\bigcap B_n$, we have $\mu(A)\le\alpha\le \mu(B)$ and $\mu(B-A)=0$.
In other words, $$\mu(A)=\alpha. $$
