Lebesgue measure of an uncountable point-set For a countable set, it is easy to construct a countable cover of arbitrary small intervals to show that the set has measure zero. Can this reasoning be extended to sets with uncountably many isolated points? By definition of the Lebesgue measure, the cover of the points must consist of finitely many intervals. 
The typical example of an uncountable set with Lebesgue measure zero is the Cantor set. However, for this set, the complement can easily be quantified and the set is constructed in countably many steps. How should uncountable point sets on which no order can be induced be dealt with?
 A: As Umberto P. already hints at, there are no uncountable subsets $E$ of $\Bbb{R}$ consisting only of isolated points; indeed, the intersection of $E$ with each $[-n,n]$ is finite (why?), so that $E$ is countable (why?). 
EDIT: As @TonyK points out, the argument above is incorrect. But it is still true that every set consisting solely of isolated points is countable. To see this, note that each $x \in E$ has an (open) neighborhood $U_x$ in $\Bbb{R}$ such that $E \cap U_x$ is finite (in fact, one can choose $U_x$ so that $E \cap U_x = \{x\}$. Now $\Bbb{R}$ is second countable and hence every subset is separable, so that each subset is Lindelöf (see e.g. Prove that a separable metric space is Lindelöf without proving it is second-countable), so that there is a countable subcover $(U_{x_n})_n$. But then $E \subset \bigcup_n [E \cap U_{x_n}]$ is countable, because each $E \cap U_{x_n}$ is finite.
Also note that the definition of Lebesgue measure requires countable covers by intervals, not finite covers (otherwise, $\Bbb{Q}$ would not have measure zero). 
An interesting set of positive (but smaller than expected) measure is obtained by
$$
\bigcup B(x_n, 1/2^n),
$$
where $(x_n)_n$ is an enumeration of $\Bbb{Q}$. 
