Lebesgue measure of intersection of sets Let $A,B\subset\mathbb{R}^n$ two Borel sets with finite and positive Lebesgue measure such that $|A|=|B|$, where $|A|$ denotes the Lebesgue measure of $A$, and such that $A\triangle B \subset\subset C(r,R)$, where $C(r,R)=B_R(0)\setminus \overline{B_r(0)}$ for fixed $3<r<R$. 
Under these assumptions, is then $|A\cap C(r,R)|=|B\cap C(r,R)|?$
Using $|A\cap C(r,R)|=|A|-|A\setminus C(r,R)|$ and $|B\cap C(r,R)|=|B|-|B\setminus C(r,R)|$, $|A\cap C(r,R)|=|B\cap C(r,R)|$ is true if and only if $|A\setminus C(r,R)|=|B\setminus  C(r,R)|$. And my guess is that the last equality holds because of $A\triangle B \subset\subset C(r,R)$ together with $|A|=|B|$. However, I don't know an exact argument.

Let $K = B_R(0) \setminus \overline{B_r(0)}$. Partition $A$ as $A = (A\setminus K) \cup (A \cap K) : = A_1 \cup A_2$. By definition $A_1\cap A_2 = \emptyset$. Similarly for $B$ we write $B = B_1 \cup B_2$. Writing the condition on Lebesgue measures of $A$ and $B$ we obtain 
$$
|A| = |A_1| + |A_2| = |B| = |B_1| + |B_2| \tag{1}
$$
and need to show $|A_2|  = |B_2|$. 
Since $A\triangle B \subset K$ we have $(A_1\cup A_2)\setminus (B_1 \cup B_2) \subset K$. But $A_2 \subset K$ by definition, hence we must have 
$$
A_1\setminus (B_1 \cup B_2) \subset K.\tag{2}
$$ Since $B_2\subset K$ and $A_1 \cap K = \emptyset$, we have $A_1\setminus B_2 = A_1$, hence $(2)$ becomes 
$$
A_1\setminus B_1 \subset K \tag{3}.
$$
Since both $A_1$ and $B_1$ have no intersection with $K$, $(3)$ implies that $A_1 \setminus B_1 = \emptyset$, i.e. $A_1\subset B_1$.
Since the above argument is symmetric with respect to $A$ and $B$, we also get $B_1\subset A_1$, i.e. $A_1 = B_1$. In particular we get $|A_1|  = |B_1|$, which together with $(1)$ shows $|A_2| = |B_2|$ and completes the proof.
