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? 
 A: 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?
To answer your question more directly, your proof is correct.  You proved correctly that the Lebesgue outer measure of any point is zero.  You used the countable subadditivity property of Lebesgue outer measure correctly.  The word "subadditivity" is used with sets (not necessarily Lebesgue measurable) that are not necessarily disjoint, and "additivity" is used with disjoint sets.  
If you have a countable, disjoint collection of Lebesgue measurable sets, then its union is Lebesgue measurable, and the Lebesgue outer measure of the union (also called its "Lebesgue measure" for short) is the sum of the Lebesgue measures of the sets in the collection.  It is true that any countable set, which includes any finite set or singleton, is Lebesgue measurable.  However, the question doesn't appear to ask anything about whether a countable set is measurable.   
