Show that any countable subset of the reals is negligible

Question

I got this task : Show that any countable subset of the reals is negligible.

Does this mean that the subset of the reals has an outer zero measure? I'm quite new too measure theory, so dont know quit where to start. Anyone got some tips?

Thanks.

Answer 1

Sketch:

• Let $X$ be the countable set and pick an order on $X$, in other words $X=\{x_1,x_2,\cdots\}$. Fix $\alpha>0$.

• For each $i$, let $I_{i,\alpha}$ be an interval of width $\frac{\alpha}{2^i}$ containing $x_i$.

• Let $I_\alpha = \bigcup I_{i,\alpha}$. Then, $\mu(I_\alpha)\leq \alpha$.

• Since $X\subseteq I_\alpha$, $\mu(X)\leq\mu(I_\alpha)\leq \alpha$. Since $\alpha$ is arbitrary, we know that $\mu(X)=0$.

Answer 2

Another approach. First prove that singletons have measure $0$. Take a singleton $\{x\}$, and note that for every $\epsilon>0$, $$ (x-\epsilon,x+\epsilon) \supset \{x\} \implies \underbrace{\mu((x-\epsilon,x+\epsilon))}_{=2\epsilon} \geq \mu(\{x\})\geq 0. $$ Since $\epsilon>0$, we conclude that $\mu(\{x\}) = 0$ for any $x$. Having proven this, let us enumerate a countable set $A$ via: $$ A = \{x_1,x_2,\dots\}. $$ Using the countable additivity of measure, we can write $$ \mu(A) = \mu\left(\bigcup_{i=1}^{\infty}\{x_i\}\right)=\sum_{i=1}^{\infty}\underbrace{\mu(\{x_i\})}_{=0} = 0. $$

Hence, any countable set has measure $0$.