Prove that there is a compact $K_{\epsilon}$, $J$-measurable s.t. $\int_{A}\chi_{A-K_{\epsilon}} < \epsilon$ 
Let $A$ be a $J$-measurable set and $\epsilon > 0$. Prove that there is a compact $K_{\epsilon} \subset A$, $J$-measurable such that
  $$\int_{A}\chi_{A\setminus K_{\epsilon}}(x)dx < \epsilon.$$

I dont know how to start this question. Informally, I think that I should a sequence of compacts $(K_{1/n})$ such that $K_{1/n} \to A$. Since $A$ is $J$-measurable, is bounded, so I need to construct a sequence of closed set in $A$ such that $(K_{1/n}) \to A$. But I dont know if it is correct neither how to formalize.
 A: Since $A$ is Jordan measurable, the boundary $\partial A$  has zero content and can be covered with a finite collection of non-overlapping closed rectangles with total volume less than $\epsilon$.  
Take an enclosing rectangle $Q \supset A$ and extend the cover of $\partial A$ into a partition $P $ of $Q$. We can decompose $P$ as a union $P = S_1 \cup S_2 \cup S_3$ of sets of rectangles $R$ where $S_1 =\{R: \, R\subset A\}$, $S_2 = \{R : \, R \cap A \neq \emptyset \}$ and $S_3 = P \setminus S_2$, and where
$$\sum_{R \in S_2} vol(R) - \sum_{R \in S_1} vol(R) < \epsilon$$
We see that $K_\epsilon = \bigcup_{R \in S_1} R$ is a compact subset of $A$ as it is a finite union of nonoverlapping, closed rectangles contained in $A$.
Since both $A$ and $K_\epsilon$ are Jordan measureable, the indicator function $\chi_A$ is Riemann integrable over those sets with 
$$\int_A \chi_A = \int_{A \setminus K_\epsilon}\chi_A + \int_{K_\epsilon} \chi_A = \int_{A }\chi_{A \setminus K_\epsilon} + \sum_{R \in S_1} vol(R),$$
and it follows that,
$$\int_{A }\chi_{A \setminus K_\epsilon} = \int_A \chi_A - \sum_{R \in S_1} vol(R) \leqslant \sum_{r \in S_2} vol(R) - \sum_{R \in S_1} vol(R) < \epsilon$$
A: Assuming you mean Jordan measurability, for any finite union of rectangles contained in $A$, there is a finite disjoint union of compact rectangles contained in them, the sum of whose volumes is close to $m(A)$, and so on.
