There is the famous isomorphism theorem for measure spaces:
All separable nonatomic normalized measure algebras are mutually isomorphic.
what about an atomic normalized measure algebras?
Is it possible to establish an isomorphism or a measure-preserving transformation from an atomic normalized measure algebra into the measure algebra of the unit interval?
These theorems would be helpful in understanding my qeustion.
Any help would be appreciated.
$\def\ed{\stackrel{\text{def}}{=}}$ Suggestion
Let $\ (\mathscr{A},\mu)\ $ be a normalised atomic measure algebra with atoms $\ a_1,a_2,\dots\ $ of weights $\ w_1\ed\mu\big(a_1\big),w_2\ed\mu\big(a_2\big),\dots $ respectively, $\ (\mathbf{T},\nu)\ $ the measure algebra of the unit interval, $\ w_0\ed0\ $, and for each $\ i\ $ $\ \alpha_i\in\mathbf{T}\ $ be the class of Lebesgue measurable sets equivalent to the interval $\ \left[\sum_\limits{j=0}^{i-1}w_j,\sum_\limits{j=0}^iw_j\right]\ $. Then $\ \nu\big(\alpha_i\big)=w_i=\mu\big(a_i\big)\ $, and it looks to me like the measure algebra generated by $\ \alpha_1, \alpha_2,\dots\ $ should be isomorphic to $\ \mathscr{A}\ $ (but I haven't done all the i-dotting and t-crossing necessary to prove it.)
Reply to query from OP in comments below
On rereading this comment, it looks like I initially misunderstood it. It now appears to me to be asking how to construct the isomorphism referred to in my answer above.
Since $\ (\mathscr{A},\mu)\ $ is an atomic normalised measure algebra, any element $\ a\ $ of $\ \mathscr{A}\ $ can be expressed uniquely as a countable join $$ a=\bigvee_{\big\{i\in\mathbb{N}\,\big|\,\mu\big(a\wedge a_i\big)>0\big\}}a_i $$ of atoms of $\ (\mathscr{A},\mu)\ $. The assignment $$ a\mapsto \bigvee_{\big\{i\in\mathbb{N}\,\big|\,\mu\big(a\wedge a_i\big)>0\big\}}\alpha_i $$ will then give an isomorphism of $\ (\mathscr{A},\mu)\ $ to the subalgebra of $\ (\mathbf{T},\nu)\ $ generated by $\ \alpha_1, \alpha_2,\dots\ $.