As per the definition; a subset $A$ of $R^n$ has ($n$-dimensional) content zero if for every $\epsilon > 0$, there is a finite cover $\{U_1,…U_n\}$ of A by closed rectangles such that $\sum_{i=1}^{n} v(U_i) < \epsilon$.
I'm rather stuck on how to show this point. In fact, aside from the idea that the measure of any finite set is also 0, I'm not very sure on techniques for showing something has content zero in generality.
My informal progress so far consists of:
Consider that the $\lim_{n\to\infty} 1/n$ = 0. So we know that all the points in the set lie inside the set [0,1]. We can reason that there does exist a finite cover of A. We need to show that there exists a finite cover where $m(S) = 0$.
An attempt I've tried only seemed to prove that the set has Lebesgue measure zero but not content zero. Consider the cover $\bigcup_{k=1}^{\infty}[\frac{1}{k}, \frac{1}{k}]$. Consider intervals $[\frac{1}{k}, \frac{1}{k}]$. Each interval $I_k$ has m($I_k = 0$) and the union completely covers the set.
Really stuck on proving it for content zero. Would appreciate any help !
Best Answer
As you correctly observed, the fact that they accumulate towards $0$ is going to be a key. Here's one way to use it in order to create a finite cover with closed intervals with total length less than any $\varepsilon>0$. The first interval will be $U_1=\left[0,\frac{\varepsilon}{2}\right]$, whose length is $\frac{\varepsilon}{2}$. Since $\lim\limits_{n\to\infty}\frac{1}{n}=0$, all but finitely many of the points of $A=\left\{\frac{1}{n}\right\}_{n=1}^{\infty}$ lie in $U_1$. So that leaves you with finitely many points $\left\{1,\frac{1}{2},\frac{1}{3},\ldots,\frac{1}{m}\right\}$, and you can cover them with closed intervals with total length less than $\frac{\varepsilon}{2}$. (Let me know if you'd like me to fill in the details for the last step.)
UPDATE. As you pointed out, we can simply make a reference to any finite set being of measure or content zero. Or we can continue with intervals: $U_2=\left[1,1+\frac{\varepsilon}{2^2}\right]$, $U_3=\left[\frac{1}{2},\frac{1}{2}+\frac{\varepsilon}{2^3}\right]$, …, $U_{m+1}=\left[\frac{1}{m},\frac{1}{m}+\frac{\varepsilon}{2^{m+1}}\right]$. If I remember my geometric sums correctly, the total length of these is $$\frac{\varepsilon}{2^2}+\frac{\varepsilon}{2^3}+\cdots+\frac{\varepsilon}{2^{m+1}}=\frac{\varepsilon}{2}\left(\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^m}\right)<\frac{\varepsilon}{2}\cdot1=\frac{\varepsilon}{2}.$$