Uncountable atomless subalgebras of the Boolean algebra of all Jordan measurable sets in [0,1] Definition: Suppose $\mathcal A$ is
the Boolean algebra of all Jordan measurable sets in $I=[0,1]$ (i.e $\mathcal A=\{A\subseteq I: \mu(\partial(A))=0\}$, where $\mu$ is the Lebesgue measure and $\partial$ means the topological boundary).
Definition: "Interval algebra" suppose $\mathcal B$ on $I=[0,1]$ is the algebra generated by the sets of the form $([a_{0},b_{0})\cup[a_{1},b_{1})\cup...\cup[a_{n},b_{n}))\cap I$ where $0\leq a_{0}< b_{0}<a_{1}<b_{1}<...<a_{n}<b_{n}$.

(1) Prove or disprove: Every (nontrivial) uncountable atomless subalgebra of $\mathcal A$ is isomorphic to $\mathcal B$.
(2) If (1) is false, I would say: Every (nontrivial) uncountable atomless subalgebra of $\mathcal A/N$ ($N$ is the family of all null sets) is isomorphic to $\mathcal B$.

By atomless I mean a Boolean algebra which has no atom.
 A: It seems to me that the cardinalities are off, so (1) fails, since $\cal A$ contains an atomless Boolean algebra of cardinality $2^{\frak c}$, and so many distinct atomless Boolean subalgebra of that size, but $\cal B$ has size continuum. So in these instances, they cannot be isomorphic.
Similarly, a cardinality argument shows at least that (2) will fail if the continuum hypothesis is false, since in that case ${\cal A}/N$ will have size continuum, and therefore by the downward Lowenheim-Skolem theorem will have atomless subalgebras of size $\aleph_1$, which will again be too small to be isomorphic to $\cal B$. So (2) fails when CH fails. 
A: An answer to (2) in ZFC:  The algebra $\mathcal B$ is not only uncountable, it  has the property that below any positive element there are uncountably many elements. 
Let  $\mathcal A'$ be the algebra of all sets of the form $([a_{0},b_{0})\cup[a_{1},b_{1})\cup...\cup[a_{n},b_{n}))\cap I$ where all $a_i,b_i$ are in $([0,\frac12]\cap \mathbb Q) \cup [\frac12,1]$.  This is a subalgebra of $\mathcal A$.  So $\mathcal A'/N$ is a subalgebra of $\mathcal A/N$, it is uncountable and atomless.
Now consider the element $[0,\frac12]/N$ in $\mathcal A'/N$.  It is positive, but there are only countably many elements below it. 
So  $\mathcal A'/N$ is not isomorphic to $\mathcal B$.
