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The question I have is the following:

Let $X: [0, \infty]\rightarrow [0, \infty]$ be a continuous function such that $X(0) = 0$. We define the "$X$-outer measure" $\mu^*$ to be $\mu^*(A) =$ inf$\{\displaystyle\sum\limits_{n=1}^\infty X(|I_n|): E\subset \cup I_n\}$ for any set $A\subset \mathbb{R}$, where the infimum is over coverings of A by countably many intervals $(I_n)_{n=1}^\infty$. Prove that $\mu^*(E) = 0$ whenever $E$ is countable.

How would I go about proving this? Any help would be greatly appreciated!

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up vote 3 down vote accepted

Since $E$ is countable, $E\subseteq\{x_n\mid n\in\mathbb N\}$. For every $n$, choose $I_n=(x_n-r_n,x_n+r_n)$, for some positive sequence $(r_n)$. Then, $E\subseteq\bigcup\limits_nI_n$ and $\sum\limits_nX(|I_n|)=\sum\limits_nX(2r_n)$. Since $X(0)=0$ and $X$ is continuous at $0$, for every positive $\varepsilon$, one can pick $r_n\gt0$ such that $X(2r_n)\leqslant\varepsilon/2^n$ for every $n$. Then $\sum\limits_nX(2r_n)\leqslant\varepsilon$, hence $\mu^*(E)\leqslant\varepsilon$. QED.

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Since X is continuous around 0, we can find a sequence of $\delta_n$'s such that $X(\delta_n)<\epsilon/2^n$. Let $E$ be countable, i.e. $E=\{x_1,x_2,...\}$. Then, take the intervals $I_n=(x_n-\delta_n/2,x_n+\delta_n/2)$. We have $\sum X(|I_n|)<\epsilon$, thus $\mu^*(E)=0$.

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