# If $\mu(E) > 0$, $\mu(F) = 0$, is $\mu(E + F) = \mu(E)$?

Let $\mu$ be standard Lebesgue measure. If $E$ is a set of positive measure and $F$ is a set of zero measure, then is it true that $\mu(E + F) = \mu(\{e + f: e \in E, f \in F\}) = \mu(E)$?

No, take $E = [0,1]$ and $F = \Bbb{Q}$.
if $E = [0;1]$ and $F = \{0;1\}$ then $E+F = [0;2]$ so $\mu(E+F) = 2 \neq 1 = \mu (E)$
If $C$ is the Cantor set then $C + C = [0,2]$.
Letting $E = C \cup D$ where $D$ has measure $1$ will result in $[0,2] \subset E + C$, so that $\mu(E + C) \geq 2$. On the other hand, $\mu(E) = 1$ since the Cantor set has measure zero.