Why is the class of all Jordan-mesurable subsets a ring?

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I have the following problem:

A bounded subset $A\subseteq \mathbb{R}^n$ is jordan mesurable if for all $\epsilon>0$ there exists some cuboids $X,Y\in \mathfrak{Q}(\mathbb{R}^n)$ s.t. $X\subseteq A\subseteq Y $ and $vol(Y)-vol(X)<\epsilon$.

I need to show that the class $\mathfrak{J}(\mathbb{R}^n)$ of all jordan mesurable subsets $A\subseteq \mathbb{R}^n$ form a ring.

My Idea was the following, where I have some troubles with the third point. They gave us also the hint to use that a bounded subset is jordan mesurable iff the boundary $\partial A$ is a lebesgue nullset, but I don't see where to use this.

I Need to show the following points:

  1. $\emptyset \in \mathfrak{J}(\mathbb{R}^n)$: Let us remark that $\emptyset \in \mathfrak{Q}(\mathbb{R}^n)$. Take $\epsilon >0$ and $X=Y=\emptyset$. Then $X\subseteq\emptyset \subseteq Y$ and $vol(X)-vol(Y)=0<\epsilon$
  2. $A,B\in \mathfrak{J}(\mathbb{R}^n)\Rightarrow A\cup B\in \mathfrak{J}(\mathbb{R}^n)$: Since $A,B\in \mathfrak{J}(\mathbb{R}^n)$ we have $X\subseteq A\subseteq Y$ s.t. $vol(X)-vol(Y)<\frac{\epsilon}{2}$ and $X'\subseteq B\subseteq Y'$ s.t. $vol(X')-vol(Y')<\frac{\epsilon}{2}$. Now remark that $X\cup X', Y\cup Y'\in \mathfrak{Q}(\Bbb{R}^n)$ since it is a ring and $X\cup X'\subseteq A\cup B \subseteq Y\cup Y'$. Furthermore let $\epsilon>0$ then $$vol(X\cup X')-vol(Y\cup Y')\leq vol(X)+vol(X')-vol(Y)-vol(Y')<\frac{2\epsilon}{2}=\epsilon$$ thus $A\cup B \in \mathfrak{J}(\mathbb{R}^n)$.
  3. $A,B\in \mathfrak{J}(\mathbb{R}^n)\Rightarrow A\setminus B\in\mathfrak{J}(\mathbb{R}^n)$:

Can someone maybe help me?

Thanks a lot