# 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.

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,

$$\int_{A }\chi_{A \setminus K_\epsilon} = \int_A \chi_A - \sum_{R \in S_1} vol(R) \leqslant \sum_{r \in S_2} vol(R) - \sum_{R \in S_1} vol(R) < \epsilon$$

• I got it! Thank you! – Lucas Corrêa Feb 21 at 3:56
• By the way, is possible to prove the question using the regularity of measures? I mean: if $A$ is Lebesgue measurable, there is a sequence $(K_{n})$ such that $K_{n} \to A$. Here, there is two points: if $A$ is Jordan measurable, then is Lebesgue measurable, or Jordan measure is regular? If is possible to use it, how can I show that each $K_{n}$ is Jordan measurable? Should I open a new question about it? – Lucas Corrêa Feb 21 at 4:02
• @LucasCorrêa: You're welcome. The brute force approach with partitions is how I learned this Riemann integration theory for $\mathbb{R}^n$. You are looking for something more elegant perhaps. I believe there is a theorem (see Munkres: Analysis on Manifolds) that may be relevant. If $A$ is open and $f$ is bounded and continuous then the extended (improper) integral exists and if the ordinary integral $\int_A f$ exists they are equal. There exists an increasing sequence of J-measurable compact sets $C_n$ such that $\cup_n C_n = A$ and $\int_{C_n} f \to \int_A f$. – RRL Feb 21 at 6:25
• This may work with $f = \chi_A$. Of course asking another question may be best. – RRL Feb 21 at 6:26
• Also what reference are you using to learn this material on multivariable Riemann integration?. Most people don't even bother to learn it assuming that Lebesgue integration is all that will be needed. – RRL Feb 21 at 6:29

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.