I know that it is possible to have deletion Cantor set which are of non-zero measure (fat Cantor sets) furthermore it must never of measure one since it would result in a contradiction (as Cantor is nowhere dense but full measure implies dense in $[0,1]$).
I have found that there exists a partition of $[0,1]$ by Cantor sets. Now if it is countable (not necessarily finite) it holds that the Lebesgue measure of the union is the sum of the measure of each set.
My question is; can we construct such union with fat Cantor sets and such that it is countable?
If so, can we "add" enough such disjoint fat Cantor sets so that the measure of the union is one (i.e. the union has full measure in $[0,1]$)?
Any help is appriciated!
Best Answer
That would violate Baire Category Theorem. So we need uncountable many fat Cantor sets to cover $[0,1]$.