It is perhaps instructive to note that there is an open subset of $\mathbb{R}$ which has finite measure and contains all rational numbers. Just number the rational numbers as $\mathbb{Q}=\{q_1,q_2,\ldots,\}$ and let $$U=\bigcup_{i=1}^\infty (q_i-2^{-i-1},q_i+2^{-i-1}).$$ The measure of $V$ is less than 1 (you can make it less than some prescribed $\epsilon$ easily enough). And the measure of open sets seems easier to understand, since any open set is the disjoint union of open intervals and the measure is the sum of the length of these intervals. (But note that the intervals defining $U$ above are not disjoint; there will be a great deal of overlap, so overlapping intervals join together to make bigger intervals …)

This is a bit counterintuitive of course, since there doesn't seem to be a whole lot of room outside of $U$, yet the complement of $U$ has infinite measure. If you think about it, it must mean that even though $U$ is the countable union of open intervals, the complement of $U$ has an uncountable number of components, each consisting of just one point. (Any connected subset of $\mathbb{R}$ is an interval, but an interval containing no rational point must be a singleton.)