Is the set {$az:z\in \mathbb Z$} for all $a \in R$ a Lebesgue null set?

Question

As far as I understand, every countable set is a Lebesgue null set.

Does that mean that the following sets are also Lebesgue null sets? In my opinion, they are all countable as we just multiply/ add a real number. Is that correct?

1. {$az:z\in \mathbb Z$} for all $a \in R$ (I'm not 100 % sure about that one, as it multiplies the z with a real number)
2. {$a+z:z\in \mathbb Z$} for all $a \in R$
3. {$a+q:q\in \mathbb Q$} for all $a \in R$
4. {$aq:q\in \mathbb Q$} for all $a \in R$
5. Every finite union of the above mentioned sets is a Lebesgue null set

Answer 1

Yes, they are countable, since they can all be seen as the image of a countable set via some mapping:

1. $\{az : z\in\mathbb Z\}=f(\mathbb Z)$, where $f:\mathbb Z\to\mathbb R$, $f(z)=az$.
2. $\{a + z : z\in\mathbb Z\}=f(\mathbb Z)$, where $f:\mathbb Z\to\mathbb R$, $f(z)=a + z$.
3. $\{a+q : q\in\mathbb Q\}=f(\mathbb Q)$, where $f:\mathbb Q\to\mathbb R$, $f(q)=a+q$.
4. $\{aq : q\in\mathbb Q\}=f(\mathbb Q)$, where $f:\mathbb Q\to\mathbb R$, $f(q)=aq$.

Since $\mathbb Z$ and $\mathbb Q$ are countable, their images via any map are too. Hence, all the above sets are countable and therefore they are Lebesgue null.

For 5, simply note that a finite union of countable sets is again countable.