lebesgue measure and countable sets [closed]

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

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MathOverflow is a site for questions and answers of interest to research mathematicians. Your question would be appropriate for a more student-oriented site. See e.g. mathoverflow.net/faq#whatnot for a list of some such sites. – Pete L. Clark Mar 13 2010 at 23:31

closed as too localized by Pete L. Clark, S. Carnahan♦, François G. Dorais♦, Qiaochu Yuan, Joel David HamkinsMar 14 2010 at 0:14

 so if the measure of a single point was not zero then measure of a countable set would be $\infty$ which would be the same as the measure of set $[0, \infty) which wouldnt make sense. – curious Mar 13 2010 at 21:44 Positive isn't enough, perhaps the measures of the points in some countable set are the terms of a convergent series. – Gerald Edgar Mar 14 2010 at 0:06 @curious: If all points had the same positive measure, then you would be correct: a countably infinite set would have infinite measure. However, there are measures in which (some) single points have positive measure, but countable sets may still have finite measure. For example, take a bijection between$f\colon\mathbb{Q}\to\mathbb{N}$, and define a measure$\mu$on$\mathcal{P}(\mathbb{R})$by letting$\mu(X) = \sum_{q\in \mathbb{Q}\cap A}2^{-f(q)}$. Then every subset of$\mathbb{R}$is measurable, and has finite measure; some points (the rationals) have positive measure, some have measure 0. – Arturo Magidin Mar 14 2010 at 20:24 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.)