Compact $C \subset$ Jordan-measurable $A$ such that $\int_{A-C} 1 < \epsilon$. This question is (3-22) in M. Spivak's Calculus on Manifolds.
If $A$ is a Jordan-measurable set and $\epsilon > 0$ show that there is a compact Jordan-measurable set $C \subset A$ such that $\int_{A-C}1 < \epsilon$.
Intuitively, I can see why this would be true. Without considering pathological examples immediately, I think: If $A$ is Jordan-measurable then it must be bounded. We could take a closed set $C \subset A$ and for each point $x$ on the boundary of $C$, reduce the distance between $x$ and the boundary of $A$ until it is sufficiently small at each point. In other words, "expand" $C$ while staying inside of $A$ until $C$ is "big enough" for our claim to hold. 
I am troubled, however, trying to prove the statement rigourously. Any help with this will be appreciated. 
 A: A definition of Jordan-measurable given in Spivak's Calculus on Manifolds is that the set is bounded, and the boundary (which is everything except the open interior and open exterior) has measure zero.
Let $B$ be the interior of the Jordan-measurable set $A$. Then $B$ is open. Cover the boundary of $A$ with open rectangles whose total volume is no more than $\epsilon$.
Let $U$ be the union of these open rectangles, so $U$ is open. Let $E$ be the exterior of $A$. Let $C$ be the complement (in the ambient space) of $E \cup U$. Then $C$ is contained in $A$. To see why, let $x \in C$ (which in fact might be empty). Certainly $x \notin E$. $x$ is also not in $U$, and so is not in the boundary of $A$, which is contained in $U$. The ambient space is partitioned into the interior, the exterior, and the boundary of $A$, so $x$ is in $B \subseteq A$.
Since $C$ is contained in $A$, it is bounded, and it is also the complement of an open set, so it's compact.
Finally, $A - C$ is contained in $U$, which is the union of the rectangles covering the boundary of $A$. This shows the Jordan measure of $A - C$ is no more than $\epsilon$. 
Detail: Because $A$ is bounded, it's easy to see that the boundary of $A$ is also bounded. The boundary of $A$ is also closed (complement of the interior and exterior). So the boundary of $A$ is compact. Then the open cover of the boundary by rectangles can be taken to be a finite cover (boundary is content zero). This means that $U$ is the union of a finite collection of open rectangles. To show that $A - C$ is Jordan-measurable, so that your integral exists, notice that $A - C = A \cap U$. Then use the fact that $Boundary(X \cup Y)$, $Boundary(X \cap Y) \subset Boundary(X) \cup Boundary(Y)$, to show the boundary of $A - C$ has measure zero.
EDIT: We have to show that $C$ is Jordan measurable.  
A set and its complement have the same boundary, so $A - C = A \cap \tilde C$ and $\tilde A \cup C$ have the same boundary.
$C = A \cap ( \tilde A \cup C)$, so $\partial C \subset \partial A \cup \partial ( \tilde A \cup C) = \partial A \cup \partial (A-C)$.
$\partial C \subset \partial A \cup \partial (A-C)$ shows that the boundary of $C$ has measure zero.
