Disjoint union of finitely many atoms

Question

I have some problems with proving this : If measure assume only finitely many values then $X$ is a disjoint union of finitely many atoms. Can someone help me ?

Answer 1

The answer is yes. Suppose $\mu$ assumes finitely many values $0 < m_1 < \cdots < m_p.$ Consider the set $\mathscr{E}$ of all measurable sets with measure $m_1;$ each element of $\mathscr{E}$ is an atom. We define a relation on $\mathscr{E}$ by setting that two of its elements $\mathrm{X}$ and $\mathrm{Y}$ are equivalent if $\mathrm{X} \triangle \mathrm{Y} := (\mathrm{X} \setminus \mathrm{Y}) \cup (\mathrm{Y} \setminus \mathrm{X})$ has measure zero (equivalently, if $\mathrm{X} \cap \mathrm{Y}$ has measure $m_1$). This creates a quotient set $\bar{\mathscr{E}}.$ For each class $\dot{\mathrm{X}}$ in $\mathscr{E}$ choose a representative $\mathrm{X} \in \dot{\mathrm{X}}$ and let $\mathscr{R}$ the set of all these chosen representatives. Observe that two elements of $\mathscr{R}$ can only intersect in a set of measure zero, also $\mathscr{R}$ is finite for if it were infinite, then it would contain a sequence $(\mathrm{R}_n)$ and each one of the representatives has measure $m_1 > 0$ and so the measure of $\cup \mathrm{R}_n$ would be $\infty.$ Since $\mathscr{R}$ is finite, write it as $\mathscr{R} = (\mathrm{R}_1, \ldots,\mathrm{R}_\nu)$ and consider $\mathrm{T}_i = \mathrm{R}_i \setminus \cup_{j < i} \mathrm{R}_j,$ this creates disjoint $\mathrm{R}_i.$ To finish off, consider $\mathrm{T}_0$ to be the complement of $\cup \mathrm{T}_i.$