I know that in general a subset of a measurable subset need not be measurable (the general example being Vitali sets). However I've been working on the following and hit a wall:
Let $U$ and $V$ be Lebesgue measurable subsets of $[0,1]$. Assume $A$ is a subset of $U$ and $B$ is a subset of $V$, $A \cup B = [0,1]$, and $m(U) + m(V) = 1$, where $m$ is the (complete) Lebesgue measure. Show that $A$ and $B$ are Lebesgue measurable.
I've found that $m(U \cap V) = 0$, and thus $A \cap B$ is Lebesgue measurable, and similarly that $A \cap V$ and $B \cap U$ are Lebesgue measurable since they're both subsets of $U \cap V$ and the Lebesgue measure is complete.
I feel like I'm missing an obvious way to express $A$ and $B$ as intersections, unions, or complements of sets that are measurable. Am I on the right track, or is their something else I'm missing?
Define $\tilde {A} = A \setminus (U\cap V) $ and define $\tilde {B},\tilde {U},\tilde {V} $ analogoulsy.
As you noted already, $U\cap V $ is a null set, so that $\tilde {U},\tilde {V} $ are measurable. Furthermore, if $\tilde {A} $ is measurable, then so is $A =\tilde {A} \cup (A \cap (U\cap V)) $, since arbitrary subsets of null-sets are null sets and thus measurable.
But we have $\tilde {A}\cup \tilde {B} = [0,1] \setminus (U \cap V) $ and $\tilde {A} \cap \tilde {B}\subset U\cap V \setminus (U\cap V)=\emptyset$.
From this, it easily follows (how?) that $\tilde {A}=\tilde {U} $ is measurable. Analogously for $\tilde {B} $.