Countable Set has Measure Zero

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Theorem

Let $S$ be a countable set.

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Then the measure of $S$ is $\map m S = 0$.

Proof

Let $\ds \set {x_i}_{i \mathop = 1}^\infty$ be an enumeration of the elements of $S$.

For any (strictly) positive real number $\epsilon$, define:

$A_i = \paren {x_i - 2^{-i} \epsilon, x_i + 2^{-i} \epsilon}$

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Then:

$\ds S \subseteq \bigcup_{i \mathop = 1}^\infty A_i$

and:

$\ds \map m {\bigcup A_i} \le \sum_{i \mathop = 1}^\infty 2^{1 - i} \epsilon = 2 \epsilon$

Since our choice of $\epsilon$ was arbitrary, for any positive real $z$ we can construct a set $T$ such that $S \subseteq T$ and $\map m T \le z$.

Hence $X$ has zero measure.

$\blacksquare$