>>13612969>>13612977Countable subadditivity shows that |A| <= |A union B|
B is not necessarily empty.
In any case I figure it out. The proof goes like this:
We cover A union B with open intervals by covering A and B separately.
Let I1, I2, ... be any cover of A
Let J1, J2, ... be a cover of B such that sum{length(Jk)} < ? (given any ?>0). We can do this since |B| = 0
C := I1, J1, I2, J2, ... covers A union B and the sum of the lengths is less than ?{length(Ik)} + ?.
This holds for any cover of A, therefore the sum of the lengths of C is less than or equal to |A| + ?.
This holds for any ?>0, therefore the sum of the lengths of C is less than or equal to |A|.
Thus, |A union B| = inf{sum of lengths of intervals: sequence of intervals that cover A union B} <= A.
