Prove that there exists $A \in F$ (σ-algebra) such that $P(A) = a$ The question
Let $\left( \Omega,F,P \right)$ be a probability space.
A set $E\in F $ is called  K-set iff
$P(E)>0$ and $B \subseteq E$ with $B \in F$ $\Rightarrow P(B) = P(E)$ or $P(B) = 0$
Note: I don't know how the above term is translated in English. I looked it up on Google but found nothing so I will name it for the sake of this post as K-set. If anyone knows the name of this term please tell me.
Suppose that the probability measure $P$ has no $K$-sets. Prove that for every $a \in [0,1]$ there exists $A \in F$ such that $P(A) = a$
Attempt
So $P$ has not $K$-sets. From this I understand that
$\forall A \in F$ with $P(A)>0$ and $\forall B \subseteq A$ with $B \in F$ we have $P(B) \neq P(A)$ and $P(B)>0$
For $a=0$, $P(\emptyset ) = 0$ and $\emptyset \in F$ since $F$ is $\sigma$-algebra.
For $a=1$, $P(\Omega) = 1$ and $\Omega \in F$ since $F$ is $\sigma$-algebra.
For $a \in (0,1)$ I thought of the idea that we can find a sequence $\{B_{n} \}$ of sets in $A$ such that $B_{n+1} \subset B_{n} \forall n $
I also tried using the Borel-Cantelli Theorem and other Propositions involving $\lim \inf B_n $ and $\lim \sup B_n $ but couldn't get to the desired result. I know that $P(B_n)$ will get smaller and smaller as $n$ gets bigger. But how do I know that $\forall a \in (0,1)$ I can find such $B_n$ such that $P(B_n) =a$ ? If this is the approach anyway.
 A: In the following, unless otherwise specified, all sets under
consideration are measurable. That is, if we say "let $A\subseteq\Omega$"
we actually mean $A\in\mathcal{F}$.

Lemma 1:
Assume that $(\Omega, \mathcal{F}, P)$ does not have $K$-set, then we have: For any $A\subseteq\Omega$ with
$P(A)>0$ and $\varepsilon>0$, there exists $B\subseteq A$ such
that $0<P(B)<\varepsilon$.
Proof of Lemma 1: Let $A\subseteq\Omega$ with $P(A)>0$ and $\varepsilon>0$
be given. Choose $B_{0}\subseteq A$ such that $0<P(B_{0})<P(A)$.
Define $A_{1}=\begin{cases}
B_{0}, & \mbox{if }P(B_{0})\leq\frac{1}{2}P(A)\\
A\setminus B_{0}, & \mbox{otherwise}
\end{cases}.$ Note that $0<P(A_{1})\leq\frac{1}{2}P(A).$ Suppose that $A\supseteq A_{1}\supseteq\ldots\supseteq A_{n}$
have been chosen. Choose $B_{n}\subseteq A_{n}$ such that $0<P(B_{n})<P(A_{n}).$
Define $A_{n+1}=\begin{cases}
B_{n}, & \mbox{if }P(B_{n})\leq\frac{1}{2}P(A_{n})\\
A_{n}\setminus B_{n}, & \mbox{otherwise}
\end{cases}.$ Clearly $0<P(A_{n+1})\leq\frac{1}{2}P(A_{n})$ and hence $P(A_{n})\rightarrow0.$
In particular, for sufficiently large $n$, $0<P(A_{n})<\varepsilon$.

We go back to your question. Let $a\in[0,1]$ be given. If $a=0$
or $a=1$, we may choose $A=\emptyset$ or $A=\Omega$ respectively.
Suppose that $a\in(0,1).$ Define an equivalence relation $\sim$ on
$\mathcal{F}$ by $A\sim B$ iff $P(A\Delta B)=0$, where $A\Delta B=(A\setminus B)\cup(B\setminus A)$.
Intuitively, $A\sim B$ means that $A=B$ modulo a $P$-null set.
If $A\in\mathcal{F}$, we denote the equivalence class containing $A$
by $\tilde{A}$. Note that if $B\in\tilde{A}$, then $P(B)=P(A)$,
so it is well-defined to specify $P(\tilde{A})=P(A)$. Define a binary
relation on $\mathcal{F}/\sim$ by $\tilde{A}\preceq\tilde{B}$ iff
$P(A\setminus B)=0$. It is routine to verify that $\preceq$ is well-defined
(i.e., if $A_{1},A_{2}\in\tilde{A}$ and $B_{1},B_{2}\in\tilde{B}$,
then $P(A_{1}\setminus B_{1})=0$ iff $P(A_{2}\setminus B_{2})=0$).
Moreover, $\preceq$ is a partial ordering on $\mathcal{F}/\sim$.
Intuitively, $\tilde{A}\preceq \tilde{B}$ means that $A\subseteq B$ modulo a $P$-null set.
Let $\mathcal{C}=\{\tilde{A}\in\mathcal{F}/\sim\,\,\,\mid\,\,\, P(\tilde{A})\leq a\}$.
Clearly $\mathcal{C}$ is non-empty. We go to show that $\mathcal{C}$
contains a $\preceq$-maximal element. Let $\mathcal{C}_{1}=\{\tilde{A}_{i}\mid i\in I\}\subseteq\mathcal{C}$
be a chain, where $I$ is an index set. Let $b=\sup_{i\in I}P(\tilde{A}_{i})$.
Note that $b\leq a$. For each $n\in\mathbb{N}$, there exists $i_{n}\in I$
such that $P(\tilde{A}_{i_{n}})>b-\frac{1}{n}.$ Define $B_{n}=\cup_{k=1}^{n}A_{i_{k}}$.
Since $\mathcal{C}_{1}$ is a chain, we can prove by induction that
$B_{n}\sim A_{i_{k}}$ for some $k\in\{1,\ldots,n\}$. Therefore $b\geq P(B_{n})\geq P(A_{i_{n}})>b-\frac{1}{n}$. Let $B=\cup_{n}B_{n}$.
By continuity of measure, we have $P(B)=b\leq a$, so $\tilde{B}\in\mathcal{C}$.
We assert that $\tilde{B}$ is an upper bound of the chain $\mathcal{C}_{1}$.
Prove by contradiction. Suppose the contrary that there exists $j\in I$
such that $P(A_{j}\setminus B)>0$. Choose $n\in\mathbb{N}$ such
that $P(A_{j}\setminus B)>\frac{1}{n}$. Since $\mathcal{C}_{1}$
is a chain, we have $\tilde{A}_{j}\preceq \tilde{A}_{i_{n}}$ or $\tilde{A}_{i_{n}}\preceq \tilde{A}_{j}$.
If $\tilde{A}_{j}\preceq \tilde{A}_{i_{n}}$, then $P(A_{j}\setminus B)\leq P(A_{i_{n}}\setminus B)=0$
which is a contradiction. If $\tilde{A}_{i_{n}}\preceq \tilde{A}_{j}$, we have
\begin{eqnarray*}
b & \geq & P(A_{j})\\
 & = & P(A_{j}\setminus B)+P(A_{j}\cap B)\\
 & > & \frac{1}{n}+P(A_{i_{n}}\cap B)\\
 & = & \frac{1}{n}+P(A_{i_{n}})\\
 & > & \frac{1}{n}+(b-\frac{1}{n})
\end{eqnarray*}
which is also a contradiction. By Zorn's lemma, $\mathcal{C}$ contains
a maximal element $\tilde{A}$.
We go to show that $P(\tilde{A})=a$ by contradiction, then it will follow that $P(A)=a$. Suppose the
contrary that $P(\tilde{A})<a$. Choose $\varepsilon>0$ such that
$P(\tilde{A})+\varepsilon<a$. Clearly $P(A^{c})>0$. By Lemma 1,
we can choose $B\subseteq A^{c}$ such that $0<P(B)<\varepsilon$.
Now $P(\widetilde{A\cup B})=P(A)+P(B)<P(A)+\varepsilon<a$. Therefore
$\widetilde{A\cup B}\in\mathcal{C}$. Clearly $\tilde{A}\preceq\widetilde{A\cup B}$,
so $\tilde{A}=\widetilde{A\cup B}$ by maximality of $\tilde{A}$.
However, $P(A\Delta(A\cup B))=P(B)>0$, contradicting to $A\sim A\cup B$.
A: For any $E\in F$ with $P(E)>0$ we can construct $E\supset E_1\supset E_2\supset\dots$ such that $P(E_n)\to 0$.
Taking their relative complements, we get $E'_n:=E\setminus E_n$ such that $P(E'_n)\to P(E)$.
Apply this construction first for $E=X$ to find $A_1:=E_n$ where $n$ is the smallest index with $P(E_n)\le a$.
Then apply it for $E=X\setminus A_1$ and define $B_1:=A_1\cup E'_m$ where $m$ is the smallest index with $P(E'_m)\ge a-P(A_1)$,
so that $A_1\subset B_1$ and $P(A_1)\le a\le P(B_1)$.
Repeat this procedure for $E=E'_m$ to construct the required sequence $A_i$.
