If $x \in \mathbb{R}$ is a Lebesgue density point for both measurable subsets $A$ and $B$, then it's also a density point for $A\cap B$.

Question

I came up with the following statement about the Lebesgue density point of two measurable sets in $\mathbb{R}$ when solving an exercise in Stein's real analysis.

$L_A := \{x:x \text{ is a point of Lebesgue density of } A\}$, and let $L_B$ be defined similarly. Then $L_A \cap L_B = L_{A\cap B}$.

My attempt to prove the statement:

The inclusion $ L_{A\cap B} \subseteq L_A \cap L_B$ is clear, by considering the following inequality $$ 1= \lim_{r\to 0} \frac{|(x-r,x+r) \cap (A\cap B)|}{2r} \leq \lim_{r\to 0} \frac{|(x-r,x+r) \cap A|}{2r} \leq 1 $$ We now prove the opposite inclusion. We assume $0$ is both a Lebesgue density point of $A$ and $B$ for simplicity. Our goal is to show that $$ \lim_{r \to 0} \frac{ (-r , r) \cap (A\cap B) }{2r } = 1. $$ Denote $A_r = A\cap (-r, r)$ and denote $B_r = B\cap (-r,r)$. Since we know that $ |A_r \cup B_r | = |A_r| +|B_r| - |A_r \cap B_r|$ by inclusion-exclusion principle, $A_r \cup B_r = (A \cup B)\cap (-r,r) \subseteq (-r,r)$, and $A_r \cap B_r = (A\cap B )\cap (-r,r)$, we have $$ 2r \geq |A_r \cup B_r| = |A_r|+|B_r| - |(A\cap B)\cap(-r,r)|\implies 1\geq \frac{|(A\cap B)\cap(-r,r)|}{2r} \geq \frac{|A_r|}{2r}+\frac{|B_r|}{2r} - 1. $$ By taking the limit $r \to 0$ and use the assumption that $0 \in L_A \cap L_B$, the RHS of the last inequality approaches $1$, and we're done.

I can't find any related statements in both Stein or other real analysis textbooks I own. This makes me to doubt about my proof. However, if the above statement is true, then we actually can use it to solve for example exercise 3 of chapter 3 in Stein.

Suppose $0$ is a point of density of measurable $E$. Show that there exists a sequence $x_n \in E$ with $x_n \neq 0$, $(x_n) \to 0$, that satisfies (a) $-x_n \in E$ for all $n$ (b) In addition, $2x_n \in E$ for all $n$.

Since this exercise can be formulated using inclusion-exclusion principle, the statement claimed above then easily solves it.

I'd like to ask whether the proof is correct. If it's false, can anyone provide a counterexample or point out the false argument that I've made. Thanks.