Are there uncountably many disjoint uncountable real null sets? It is easy to think of a countable family of disjoint Cantor sets, and their union is of course a null set. 
It is equally trivial to define an uncountable family of Cantor sets but, how can it be ensured that they are pairwise disjoint? 
Would their union be a null set?

First do it for $C=\{0,1\}^{\mathbb{N}}$. For every $b=(b_n)_n\in \{0,1\}^{\mathbb{N}}$ let $C_b=\{(a_n)\ \mid\ a_{2n} = b_n \textrm{ for all } n\}$. Clearly $(C_b)_b$ form a partition of $C$ into compact subset of measure $0$.
Now $[0,1]$ is measure equivalent to $C$ under the map 
$a\mapsto s(a) = \sum a_n/2^{n+1}$. We can eliminate the countable subset of "undetermination" of $s^{-1}$ (the binary fractions) by considering the images $s(C_b)$, where $b$ is a sequence that is not eventually constant. We get in this way an uncountable family $(s(C_b))_b$ of compact disjoint subsets of $[0,1]$ of measure $0$. The complement of their union in $[0,1]$ is again uncountable and of measure $0$. 
Note that for $b$ not an eventually constant sequence we can describe $s(C_b)$ to be the set of those real numbers in $[0,1]$ such that their binary expansion $\sum a_n/2^{n+1}$ has $a_{2n}= b_n$   ( prescribed digits at odd positions after the binary dot). 
While at it, it should be clear how to partition each $s(C_b)$ ($b$ not eventually constant) into an uncountable family of Cantor subsets. 
