# A countable set has outer measure zero. Explanation of Royden example.

This is an example from Royden's Real Analysis 4th edition that I am having a lot of trouble understanding.

A countable set has outer measure zero. Let $C$ be a countable set enumerated as $C=\{c_k\}_{k=1}^{\infty}$. Let $\epsilon > 0$. For each natural number $k$, define $I_k=(c_k-\epsilon/2^{k+1}, c_k+\epsilon/2^{k+1})$. The countable collection of open intervals $\{I_k\}_{\infty}^{k=1}$ covers $C$. Therefore:

$$0 \leq m^*(C)\leq \sum^{\infty}_{k=1} \ell(I_k)=\sum^{\infty}_{k=1}\epsilon/2^k = \epsilon$$

This inequality holds for each $\epsilon > 0$. Hence $m^*(E) = 0$.

What I understand along with questions:

• A countable set is a set like the natural numbers.
• $C=\{c_k\}_{k=1}^{\infty}$ is just a way to describe said countable set using indexing for all of the elements in the set.
• $I_k=(c_k-\epsilon/2^{k+1}, c_k+\epsilon/2^{k+1})$ I kind of understand. This is defining an open cover over the countable set (I think).
• Why the $\epsilon/2^{k+1}$ instead of just $\epsilon$? It looks like the interval is continually shrinking as n goes to infinity.
• The equation in the middle is summing up all of the intervals. The outer measure is simply the sum of all of the open intervals in the set C. $m^*(C)$ is countably subadditive, hence the $\leq$ sign here: $m^*(C)\leq \sum^{\infty}_{k=1} \ell(I_k)$
• How is the last summation producing epsilon?
• How is this summing to zero?

-Idle

• Why $\epsilon /2^{k+1}$ rather than $\epsilon$? $\sum_{k=1}^\infty\epsilon$ diverges, which isn't very useful. Jan 17 '18 at 6:14

6) $\displaystyle\sum_{k=1}^{\infty}\frac{\epsilon}{2^{k}}=\epsilon\sum_{k=1}^{\infty}\dfrac{1}{2^{k}}=\epsilon\cdot\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}=\epsilon$.
7) No, the sum is $\epsilon$, but now we have $0\leq m^{\ast}(C)<\epsilon$ for every $\epsilon>0$, if $m^{\ast}(C)>0$, then put $\epsilon$ to be $m^{\ast}(C)$, then we arrive to $m^{\ast}(C)<m^{\ast}(C)$, a contradiction.