(1) Existence of sets of small measure.
Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$. It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$.
(2) The space and any measurable set is a numerable union of set of small measure.
Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup_n A_n$ union of a numerable set of measurable sets of measure $\mu(A_n)\le \varepsilon$
Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$. There is a sequence $(M_n)$ of sets in $\mathcal M$ with $\lim_n\mu(M_n)=\lambda$. Then $M_0:=\bigcup M_n\in \mathcal M$ (by definition) and $\lambda(M_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M_0$.
(3) Construct a set of measure $a$.
Now given $0<a<\mu(\Omega)$ we construct an increasing sequence $(A_n)$ of measurable sets with $0<a-\mu(A_n)<\varepsilon_n$ for any sequence of strictly decreasing positive numbers with $\varepsilon_n\to0$ and $\varepsilon_1<a$
Since $\Omega=\bigcup_n M_n$ is a union of measurable sets of measure $\le \varepsilon_1$, we may take $A_1=\bigcup_{n=1}^N M_n$, where $N$ is the greatest integer such that the measure of the union is less than or equal $a$. To construct $A_2$ consider $\Omega\smallsetminus A_1$ and construct $B_2\subset \Omega\smallsetminus A_1$ with $(a-\mu(A_1))-\mu(B_2)\le\varepsilon_2-\varepsilon_1$ and put $A_2=A_1\cup B_2$. The construction follows in the same way from this point.
Notice that the construction of $B_2$ and that of $A_1$ are totally similar.
