Measure of sums of sets Is it true that 
$$m(A+B)\geq m(A)+m(B)$$? For any measurable sets $A,B$
where $A+B=\{x+y,x\in A,\,y\in B\}$
Thanks 
 A: In general not; just consider e.g. the Dirac measure $\delta_1$ and the sets $$A := \{1\}, \qquad B := -A = \{-1\}.$$
A: @mathmath,
This is true for $A,B \in \mathbb{R}$ IF $A+B$ is also measurable. In general n-dimensional Euclidean space, the inequality looks a little different, i.e. if we define $A,B \subset \mathbb{R}^{n}$, then indeed $m(A+B)^{1/n} \geq m(A)^{1/n} + m(B)^{1/n}$. The only sticking point here is that, even though both $A$ and $B$ may be measurable, $A+B$ is not necessarily measurable. If $A,B, A+B$ are each measurable, then the inequality in fact holds.
There are many proofs of this available online and in print, my favorite (and I believe clearest) of which from the Princeton Lecture Series proceed as follows: (1) show the inequality is true when $A,B$ are rectangles with side lengths $a,b$ respectively. (2) show inductively the inequality is true when $A,B$ are the union of finitely many rectangles. (3) extend to the case where $A,B$ are arbitrary compact sets (which implies that $A+B$ is also compact), then (4) approximating $A,B$ as measurable sets by using compact sets instead of rectangles...
