Any countable set has measurable zero

Question

To demonstrate that ANY countable set has measure zero, is it sufficient to show that the natural numbers have a measure zero? If so, why; and, if not, why not?

Thank you :)

Answer 1

(Asuuming we are talking about Lebesgue measur)

Let $a_i$ be a sequence of a countable set $A$, then $A_i=(a_i-2^{-i-1}\varepsilon,a_i+2^{-i-1}\varepsilon)$.

Now $A\subseteq \bigcup A_i$, and $\mu\left(\bigcup A_i\right)\le\sum2^{-i}\varepsilon=\varepsilon$

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Showing that $\Bbb N$ is not enough because measure does not preserve by bijection, a simple example is the cantor set and the interval [0,1], where they have the same cardinality(hence there exists bijction between them) but one is measure 0 and the other 1.