Prove that a countable subset of $\mathbb{R}$ has Lebesgue outer measure zero

Prove that a countable subset of $\mathbb{R}$ has Lebesgue outer measure zero.
I believe I am on the right track but can use some help with this one. Here is my proof thus far:

Let $a \in \mathbb{R}$. Then, $\{a\} \subset [a-\epsilon, a+\epsilon]$ holds $\forall \epsilon>0$ and so $\lambda^*(\{a\}) \le \lambda^*([a-\epsilon, a+\epsilon])=2\epsilon \; \forall \epsilon >0$. Therefore, $\lambda^*(\{a\})=0$ holds $\forall a \in \mathbb{R}$. If $A = \{ a_1, a_2,... \} = \bigcup_{n=1}^{\infty} \{a_n\}$ is a countable set, then note that $\lambda^*(A) \le \sum_{n=1}^{\infty} \lambda^*(\{a_n\})=0$ so that $\lambda^*(A)=0$.

Is this along the right lines?

• Yes, if you had seen that Lebesgue measure is countably additive, then proving that a singleton set has measure $0$ (and your proof is basically correct) implies that any countable set has measure $0$, by countable additivity. – Ittay Weiss Dec 8 '13 at 23:46
• @IttayWeiss The Lebesgue measure of an interval its length. So, can we just use the fact that $\{a\} = [a,a]$ to prove that $\lambda^*(\{a\}) = \lambda^*([a,a]) = a-a=0$? – user193319 Oct 25 '16 at 20:56

You can use the basic definition of Lebesgue outer measure. Let $\epsilon>0$. Assume for simplicity that $R$ is countably infinite, with $R=\{r_1, r_2, r_3, \ldots\}$. For each $n \in \mathbb{N}^+$, find an interval (you can use either closed or open) with length $\epsilon/2^n$ containing $r_n$. What's the sum of the lengths of all the intervals? What can you conclude from this?