**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 $0\leq a\_{0}< b\_{0}<a\_{1}<b\_{1}<...<a\_{n}<b\_{n}$.



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> (1) Prove or disprove: Every (nontrivial) uncountable atomless subalgebra of $\mathcal A$ is isomorphic to $\mathcal B$.
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> (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 $\mathcal B$.
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By atomless I mean a Boolean algebra which has no atom.



Uncountable atomless subalgebras of the Boolean algebra of all Jordan measurable sets in [0,1]
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. 


