Non-atomicity of the uniform measure space on $(0, 1]$.

Question

I was working on the following exercise:

Show that $((0,1], \mathcal{B}_{(0, 1]}, U)$ defines a non-atomic measure space.

Above, $\mathcal{B}$ denotes the Borel $\sigma$-algebra, and $U$ denotes Lebesgue measure restricted to this sub-$\sigma$-algebra of $\mathcal{B}_{\mathbf{R}}$. Importantly, all I know about $U$ (for the purposes of this problem) is that (1) it is a measure, and (2) it is the unique measure such that if $A = \cup_{n=1}^N (a_n, b_n] \subset (0, 1]$, where $N \in \mathbf{N}$ and the intervals are disjoint, then $U(A) = \sum_{n=1}^N (b_n - a_n)$. [In words, "$U$ is finitely-additive over disjoint, finite unions of half-open sub-intervals of $(0, 1]$"]

In working on the exercise, I thought it would be useful to try to prove a property about the measure on the entire $\mathcal{B}_{(0, 1]}$. Suppose in fact it is true that for any measurable $A$,

$$ U(A) = \sup\left\{ U(I) : I \subseteq A, \text{$I$ is a disjoint finite union of half-open sub-intervals of $(0, 1]$}\right\} $$

If this statement is true, then the claim follows: if I set $\epsilon > 0$, then I get the following approximation result: $$ \exists A \supseteq I = \cup_{n=1}^N (a_n, b_n] \text{ (disj.)}: \qquad U(A) - \epsilon < U(I) = \sum_{n=1}^N (b_n - a_n). $$ Suppose now that $U(A) > 0$. If I take $\epsilon = U(A)/2$, then this says there exists a set $C = \cup_{n=1}^N (a_n, b_n] \subset A$, such that $U(C) > U(A)/2$. Additionally $U(C) \leq U(A)$ since $C \subseteq A$. Now set $B = \cup_{n=1}^N (a_n/2, b_n/2]$. Then $B \subsetneq A$, and since everything has positive measure, the strict inequalities $$ O < U(B) < U(A) $$ follow.

So the questions I have are

1. Am I on the right track here? Maybe this supremum approximation argument is overkill.
2. If this seems like a good approach, I'm struggling a bit to show that the equality holds. In particular if we fix $A$ measurable, and let $S$ denote the RHS (the supremum), then the inequality $U(A) \geq S$ is immediate, since the sup is occurring over sets $B$ all of which satisfy $U(A) \geq U(B)$ as $B \subseteq A$. So the real part that I could use help with is demonstrating $U(A) \leq S$. Again if this seems like the right thing to look at, I just want a hint.

Answer 1

Suppose for a contradiction that the set $A$ would be an atom. We may partition $(0,1]$ into intervals

$$ B_{k,n} := [(k-1)/n,k/n] ~~~(1 \le k \le n) $$

for $n = 1, 2, \ldots$. Suppose that $\mu(A) = \epsilon$. Then choose $n$ sufficiently small so that $1/n < \epsilon$ (Achimedean axiom) and observe that one of the sets $$ B_{k, n} \cap A $$ has positive measure (otherwise $\mu(A) = 0$), but also measure less than $\epsilon$.

Answer 2

You are not on the right track.

For instance observe that the supremum takes value $0$ on a sets like $(0,1]-\mathbb Q$ and also on sets like $\mathbb Q$.

For $u\in(0,1]$ and $\epsilon>0$ small enough we have $u\in(u-\epsilon,u]\subseteq(0,1]$ implying that: $$U(\{u\})\leq U((u-\epsilon,u])=\epsilon$$

Since we can take $\epsilon>0$ as small as we like this can only be true if:$$U(\{u\})=0$$