The Lebesgue Measure of the Cantor Set Why is it that m(C+C) = 2, where C is the Cantor set, but C is itself measure zero?  What is happening that makes this change occur?  Is it true generally that m(A+B) =/= m(A) + m(B)?  Are there cases in which this necessarily holds?  I should note that this is all within the usual topology on R.  Thanks!
 A: Adding two sets $A$ and $B$ is in general much more than taking the union of a copy of $A$ and a copy of $B$. 
Rather, $A+B$ can be seen as made up of many slices that look like $A$ (one for each $b\in B$), and the Cantor set has the same size as the reals, so here we have in fact as many slices as real numbers. Therein lies the problem: Even $m(A+\{1,2\})$ may be larger than $m(A)+m(\{1,2\})$, and it can be as large as $2m(A)$ (think, for example, of $A=[0,1]$). -- Of course, the slices do not need to be disjoint in general, but this is another issue.
Ittay's answer indicates that if $A,B$ are countable we have $m(A+B)=m(A)+m(B)$, more or less for trivial reasons. There does not seem to be a natural way of generalizing this to the uncountable, in part because Lebesgue measure can never be $\mathfrak c$-additive, meaning that even if we have that $A_r$ is a measurable set for each $r\in\mathbb R$, and it happens that $\sum_r m(A_r)$ converges, and it happens that the $A_r$ are pairwise disjoint, in general $m(\bigcup_r A_r)$ and $\sum_r m(A_r)$ are different. For example, consider $A_r=\{r\}$.
A: Yes, in general $m(A+B) \ne m(A)+m(B)$.  Equality does hold when $m(A+B)=0$, but for trivial reasons.
A: When $m(A)=m(B)=0$ and $A$ is countable then $m(A+B)=0$. This is so because $A+B=\bigcup _{a\in A}a+B$ is a countable union, and $m$ is translation invariant so $m(a+B)=m(B)=0$, so countable (sub)additivity gives that $m(A+B)=0$. 
However, when both $A$ and $B$ are uncountable, this argument fails, as then $A+B=\bigcup _{a\in A}a+B$ is still true but this is not a countable union. The Cantor set illustrates that the measure of $A+B$ can then indeed increase.
