A question about the range of a positive measure It is a well-known fact that the range of a positive measure $\mu$ on a measure space $(X,\cal M)$ is the interval $[0,\mu(X)]$ provided $\mu$ is atomless (i.e., there is no measurable set $A\in \cal M$ such that $\mu(A)>0$ and for any measurable set $B\subset A$ we have $\mu(B)=0$ or $\mu(B)=\mu(A)$).
It seems that the 'standard' proof of this fact involves some transfinite induction argument (cf. e.g. Halmos 1947, Wikipedia or else), but I am looking for a proof not using transfinite induction as suggested by some exercises in Bourbaki or Dieudonné Analysis II.
This would entail proving the following:
Let $r\in ]0,\mu(X)[$, and let $\cal C=\{A\in \cal M:\mu(A)\le r\}$. Suppose that $\mu$ has a 'gap' at $r$, meaning that we have $\alpha<r$, where $\alpha=\sup\{\mu(A):A\in \cal C\}$. Then $\mu$ must have an atom.
A simple path leading to that result would be to prove that in the presence of a gap at $r$, $\cal C$ is closed under finite union; in other words, if $A,B$ are such that $\mu(A)\le \alpha$ and $\mu(B)\le \alpha$, then $A$ and $B$ have no other choice but to overlap in such a way that $\mu(A\cup B)\le \alpha$. Equivalently if $A$ and $B$ are supposed also disjoint, we must have $\mu(B)\le \alpha-\mu(A)$.
The intuitive argument is that if neither $A$ nor $B$ contains an atom, then $A\cup B$ cannot contain it. I would suspect the proof to be well-known, but my brain seems unable to produce it.
 A: (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) Given a measurable set $M$ and two numbers $0<a<b<\mu(M)$, there exists a measurable set $A\subset M$ with $a<\mu(A)<b$.
Let $\varepsilon=(b-a)/2$, by (2) the set $M$ can be put as a union of measurable sets of measure less than $\varepsilon$,  $M=\bigcup_n M_n$ with $\mu(M_n)\le \varepsilon$.  Then $$\lim_N\mu\Bigl(\bigcup_{n=1}^N M_n\Bigr)=\mu(M).$$ Therefore there is some $N$ such that $a<\mu\Bigl(\bigcup_{n=1}^N M_n\Bigr)<b$, since each $M_n$ have measure $\le (b-a)/2$.
(4) Construct the set with measure $\mu(A)=a$ for a given $0<a<\mu(\Omega)$.
We will construct a sequence of measurable sets $(A_n)$ with $A_n\le A_{n+1}$ and such that
$$\frac{a+\mu(A_n)}{2}<\mu(A_{n+1})< a.$$
It is clear that in this case $A=\bigcup_n A_n$ have measure $a$.
Start with $A_0=\emptyset$. We must construct $A_1$ with $a/2<\mu(A_1)<a$. Sincer $\Omega$ have measure $>a$, we have constructed such a set in (3).
Assume we have constructed $A_n$, then $\mu(A_n)<a$, applying (3) we construct $B_n\subset \Omega\smallsetminus A_n$ whose measure is greater than half the difference with our objective:  $a-\mu(A_n)>\mu(B_n)>\frac{a-\mu(A_n)}{2}$.
Hence  $A_{n+1}=A_n\cup B_n$ satisfies our requirement. First $A_n$ and $B_n$ are disjoint $\mu(A_{n+1})=\mu(A_n)+\mu(B_n)<a$, and $\mu(A_n)+\mu(B_n)>\frac{a+\mu(A_n)}{2}$.
A: As note 'added in proof': the above answers in the positive the point I was trying to make, i.e. that in the presence of a gap at some $r>0$, the measure $\mu$ cannot be atomless.
Indeed, suppose that $\mu$ has a gap at $r$. Let $\alpha=\sup\{\mu(A):A\in \cal C\}$ and $\beta=\inf\{\mu(A):A\in \cal M, \mu(A)\ge r\}$. If $\mu$ were atomless, we know that for any $\epsilon>0$, we could write $X=\cup_{n\ge 1}A_n$ where the $A_n$'s are measurable sets of measure $\le \epsilon$. Hence for some $N\ge 1$ we would have $\alpha< \mu(\cup_{n=1}^{N} A_n)\le \alpha+\epsilon$, and taking $\epsilon<\beta-\alpha$ would yield a contradiction. Hence $\mu$ must have an atom.
