Suppose that $A$ has Jordan content null. Show that $\partial A$ has Jordan content null. 
Denote the Jordan content $\left(\displaystyle \int_{R}\chi_{A}(x)\mathrm{d}x\right)$ of $A$ by $c(A)$ and the $m(A)$ the Lebesgue measure of $A$.
(a) Suppose that $c(A) = 0$. Show that $c(\partial A) = 0$.
(b) Suppose that $m(A) = 0$. What about $m(\partial A)$?


I'm confused. I thought the content of Jordan was only defined for $J$-measurable sets. For $J$-measurable sets, the initials hypotheses seems unnecessary. I proved the following results:

Lemma 1. A bounded set $X \subset \mathbb{R}^{m}$ is $J$-measurable iff $m(\partial X) = 0$.
Lemma 2. If $K$ is compact then $c(K) = 0$ iff $m(K) = 0$.

By lemma 1 and lemma 2 we have the following equivalences:

$X$ is $J$-measurable $\Longleftrightarrow$ $m(\partial X) = 0$ $\Longleftrightarrow$ $c(\partial X) = 0$,

since if $X$ is bounded, $\partial X$ is compact.
So, if $A$ is $J$-measurable $c(A)$ and $m(A)$ are irrelevants.

Thus, I suppose that we can define $c(A)$ for arbitrary sets or I'm confused about the definitions and lemmas. Can someone help me?
 A: For part (b) note that with $A = [0,1] \cap \mathbb{Q}$ we have $m(A) = 0$ but $m(\partial A) = m([0,1]) = 1$.
For part (a) it might be helpful to recall the elementary definition of Jordan measureability and content.  
For $A \subset \mathbb{R}^m$, let $\mathcal{R}_*(A)$ denote the collection of simple sets (finite unions of nonoverlapping rectangles) that are contained in A, and let $\mathcal{R}^*(A)$ denote the collection of simple sets  that cover A.  
We define the inner and outer Jordan content,  $c_*(A)$ and $c^*(A)$, respectively, as 
$$c_*(A) = \sup_{\cup_k R_k \in \mathcal{R}_*(A)} \,\,\sum_{k=1}^n vol(R_k), \\ c^*(A) = \inf_{\cup_k R_k \in \mathcal{R}^*(A) } \,\,\sum_{k=1}^n vol(R_k)$$
If $c_*(A) = c^*(A) = c(A)$, then the set A is said to be Jordan measurable with content $c(A)$.
To connect with your lemmas, the Riemann integral of $\chi_A$ over a rectangle $R \supset A$ exists if and only if the measure of the boundary $\partial A$ is zero since the indicator is discontinuous on $\partial A$. Next, we have that $A$ is Jordan measurable if and only $\chi_A$ is Riemann integrable, and this is shown by associating lower and upper Darboux sums with the unions of rectangles used to define inner and outer content.   In this case, independent of the choice for $R$,
$$c(A) = \int_R\chi_A$$
When you are starting with the hypothesis that $A$ has content null and you take this to mean also that $A$ is Jordan measurable, then, of course, it follows from the preceeding that $m(\partial A) = 0$.
On the other hand, the notion of content null is often introduced without reference to Jordan measurability and the nature of the boundary.  The definition is that $A$ has content null if for any $\epsilon > 0$ there exists a finite collection of rectangles $\{R'_k\}_{k=1}^n$ such that
$$A \subset \bigcup_{k=1}^n R'_k, \,\,\,\sum_{k=1}^nvol(R'_k) < \epsilon,$$
and, without loss of generality, we can assume these rectangles are not overlapping.
Going back to the elementary definition of content we see that if $A$ has content null, then
$$0 \leqslant c_*(A) \leqslant c^*(A) = \inf_{\cup_k R_k \in \mathcal{R}^*(A) } \,\,\sum_{k=1}^n vol(R_k) \leqslant \sum_{k=1}^nvol(R'_k) < \epsilon$$
Since this is true for any $\epsilon > 0$, it follows that $A$ is Jordan measurable with $c(A) = 0$  - and as shown above we must have $m(\partial A)= 0$.  Then following your logic, $\partial A$ is compact and this implies $c(\partial A) = 0$.
A: Here is a lazy proof using the fact that $f$ is Riemann integrable iff $f$ is bounded, has bounded support and the set of discontinuities has measure zero.
$A$ is Jordan measurable iff $1_A$ is Riemann integrable. As above, let $c(A)= \int_A 1_A$.
(Note that $1_A$ being Riemann integrable implies that $A$ is bounded.)
Suppose $A$ is any Jordan measurable set. Then $1_A$ has bounded support and the set of discontinuities of $1_A$ has measure zero. Note that $x$ is a point of discontinuity of $1_A$ if $x \in \partial A$. Hence $\partial A$ has measure zero.
Since $\partial A$ is closed and bounded, it is compact and hence has Jordan content zero, that is $c(\partial A) = 0$.
Suppose $D$ is a countable set that is dense in $[0,1]$. Then $m(D)= 0$ since $D$
is countable, but $\partial D = [0,1]$ and so $m(\partial D) = 1$.
