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.
Let $\epsilon > 0$ be given. The integrability of $f$ allows to choose a partition $P$ of $[a,b]$ such that
$$ S_Pf - s_Pf < \epsilon $$
Say $I$ is a subinterval determined by the partition $P$ and write
$$ S_Pf - s_Pf = \sum_I ( \sup_{x \in I} f( x) - \inf_{x \in I} f(x) ) Lentgh(I) $$
It is easy to see that (Denote $ \Gamma_f = \{ (x, f(x) ) \in \mathbb{R} \times \mathbb{R} : x \in [a,b] \} $ ) that
$$ \Gamma_f \subset I \times [\inf_{x \in I} f, \sup_{x \in I} f] = \mathcal{C}$$
Hence, $\mathcal{C}$ provides a cover for the graph of $f$ and finally
$$ \sum Area ( \mathcal{C} ) = \sum_I ( \sup_{x \in I} f( x) - \inf_{x \in I} f(x) ) Lentgh(I) < \epsilon$$
Best Answer
This is not true. For example, consider the set of rational numbers $\mathbb{Q}$, which has measure zero but not content zero. To see this set has measure zero, consider the following $\mathbb{Q} \subset \cup^{\infty}_{i = 1} \{A_i\}$, where $A_i = (x_i - \frac{\epsilon}{2^i+1}, x_i + \frac{\epsilon}{2^i+1})$, and each $x_i$ is a rational number. As the rational numbers are countable, that means we can be sure that this infinite union covers $\mathbb{Q}$. The volume of $A_i$ is $v(A_i) = \frac{2\epsilon}{2^{i+1}} = \frac{\epsilon}{2}$. Then $v(\mathbb{Q}) \leq \sum ^{\infty}_{i=1}v(A_i) = \sum ^{\infty}_{i=1}\frac{\epsilon}{2^i} = \epsilon\frac{1}{1-{\frac{1}{2}}} = 2\epsilon, \forall\epsilon$. Thus we have found a cover of \mathbb{Q}, whose volume, $v(Q) = 2\epsilon$, can be made arbitrarily small.
However, there is no way we can find a finite number of sets whose union contains $\mathbb{Q}$, and whose volume is zero. Thus, $\mathbb{Q}$, does not have content zero.