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 question about the range of a positive measure
(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 $\mu(A\_n)+\mu(B\_n)>\frac{a+\mu(A\_n)}{2}$.


