Question about Jordan Measure I'm using the term "content" for Jordan measure and the term "measure" for Lebesgue measure.
The definition of content that I was given is:

A bounded set $D⊆ℝ^n$ has content if $x ↦ 1$ is Riemann integrable on $D$ and in that case we say that $vol(D) = ∫_D1$ is the content of $D$.

I know (without proof) that $D$ has content if and only if $D$ is bounded and $∂D$ has measure zero.
Now if $∂D$ has content zero, then $D$ has content (because a zero content set is also a zero measure set). However it should be false the if $D$ has content then $∂D$ has content zero. So the following proof of this statement must contain a mistake. Can someone help me find it?

If $D$ has content then there is an interval $I⊇D$ such that $vol(D) = ∫_Dχ_D$ exists (and we can assume $ClD ⊆ IntI$). This means that for each $ε > 0$ there is a partition $P$ of $I$ such that $U(χ_D,P) - L(χ_D,P) < ε$. I'm denoting with $J$ the generic subinterval of $I$ determined by $P$. Let $C$ be the set of those $J$s which contain both points of $D$ and points of $I-D$, and C' the set of those $J$s which aren't in C. If $x ∈ ∂D$ then $x ∈ IntI$, so we can find a $J$ containing both $x$ and a point of $D$ and a $J$ containing both $x$ and a point of $I-D$. It follows that $∂D ⊆ ∪C$. But
  $$ ε > U(χ_D,P) - L(χ_D,P) = ∑_J osc(χ_D,J)vol(J) = ∑_{J∈C} osc(χ_D,J)vol(J) + ∑_{J∈C'} osc(χ_D,J)vol(J) = ∑_{J∈C} vol(J).$$

 A: Actually if $D \subset I$ is bounded and $vol(D) = \int_I\chi_D$ exists then the boundary $\partial D$ has content zero.
The closure $\bar{D} = D \cup \partial D$ is closed and bounded and, hence, compact. Let $\mathcal{O} = \{O_n\}$ be an open cover of $\partial D.$ Then $\mathcal{O'}=\mathcal{O} \cup \{int(D)\}$ is an open cover of $\bar{D}$. Since $\bar{D}$ is compact there exists a finite subcollection $\mathcal{G} =\{G_1,G_2,\ldots, G_n\} \subset \mathcal{O'}$ which covers $\bar{D}$ as well as $\partial D \subset \bar{D}$. If $int(D) \notin \mathcal{G}$ then we have found an open subcover $\mathcal{G} \subset \mathcal{O}$ of $\partial D$. If $int(D) \in \mathcal{G}$ then one of the sets $G_j = int(D)$ and $\{G_k : k \neq j\} \subset \mathcal{O}$ is an open subcover of $\partial D$.  Therefore, $\partial D$ is compact.
Since $\chi_D$ is integrable over $I$ and discontinuous on $\partial D$, we must have $\partial D$ of measure zero.  Since $\partial D$ is compact it must also be of content zero.
