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 (hence $\cal C$ is a sub-$\sigma$-algebra of $\cal M$); 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.

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.

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 (hence $\cal C$ is a sub-$\sigma$-algebra of $\cal M$); 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 $B$ contain 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.

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 (hence $\cal C$ is a sub-$\sigma$-algebra of $\cal M$); 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.