Question

the lebesgue integral $ \int_{[0,1]} 1_{\mathbb{Q}} dm = 0 $ . and if we integrate the complement $ \int_{[0,1]} 1_{\mathbb{Q}^C} dm =1 $ which is the same as $\int_{[0,1]} dm $ to me this is still a bit of a mystery to there are $\infty $ rationals in the interval $[0,1] $ yet if we exclude all these rational the measure is still the same as if we included the set $\mathbb{Q} \cap [0,1] $ . why is that, i know the measure of a countable set is zero, but why i cant find an explanation for this, but how can a set with cardinality of $\infty$ still be zero

Answer 1

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.)

Answer 2

A single point has measure zero. A countable set is the countable union of points, and since the measure is countably additive, you have that the measure is the sum of the measure of the single points.