Prove that set has zero Jordan content iff its closure has measure 0 Prove that set has zero Jordan content iff its closure has measure 0.
I am having trouble with both directions , any tips would be great. THanks!
 A: I will use $m$ to denote the Jordan measure and $\lambda$ to denote the Lebesgue measure. Note that for rectangles $R$, we have $m R = \lambda R$. 
I believe that the statement should be for bounded sets. If a set has Jordan Content zero, then it is automatically bounded. It follows that any unbounded set cannot have Jordan Content zero. For example, $\mathbb{Z}$ is closed and has Lebesgue measure zero, but cannot have Jordan Content zero.
Hence I will assume that we are talking about bounded sets.
Suppose $A$ has Jordan Content zero. Then for any $\epsilon>0$, there is a finite collection of rectangles $R_i$ such that $A \subset \cup_i R_i$ and $\sum_i m R_i < \epsilon$. Since $m R_i = m \overline{ R_i}$, we can take the rectangle to be closed, hence $C = \cup_i \overline{ R_i}$ is closed. In particular $\overline{A} \subset C$.
Now let $\epsilon = \frac{1}{n}$, and let $C_n = \cup_i \overline{R_i^{(n)}}$ be the corresponding closed set, then $\overline{A} \subset C_n$ and $\sum_i m R_i^{(n)} = \sum_i \lambda R_i^{(n)}  < \frac{1}{n}$.
Then we have $\overline{A} \subset \cap_n C_n$, and so $\lambda \overline{A} \le \lambda C_n \le \sum_i \lambda R_i^{(n)}  < \frac{1}{n}$ for all $n$. Hence $\lambda \overline{A} =0$.
Now suppose $\lambda \overline{A} = 0$. There is no loss of generality in assuming that $A$ is closed (since a cover of the closure certainly is a cover of the original set). Since $A$ is bounded, we see that $A$ is, in fact, compact.
I need to show that for any $\epsilon>0$, I can cover $A$ by a finite collection of rectangles $R_i$ such that $\sum_i m R_i < \epsilon$. Since $\lambda A = 0$, there is a countable collection of rectangles $R_i$ such that $A \subset \cup_i R_i$ and $\sum \lambda R_i < \frac{1}{2}\epsilon$. For each $i$, we can find an open rectangle $O_i$ such that $R_i \subset O_i$ and $\lambda O_i \le \lambda R_i + \frac{1}{2^{i+2}} \epsilon$. Hence $\sum \lambda O_i < \epsilon$. Since $A$ is compact, and $O_i$ form an open cover, there is a finite subcover $O_{k_1},...,O_{k_m}$, and clearly $\sum_{i=1}^m m O_{k_i}  = \sum_{i=1}^m \lambda O_{k_i} \le  \sum_i \lambda O_i < \epsilon$. Hence $A$ has Jordan content zero.
