Is the sum (difference) of Borel set with itself a Borel set?

Let $d \in \mathbb{N}$, $A \subset \mathbb{R}^d$, be a Borel set. Consider Minkowski sums $$ \mathbb{S}(A) = A + A = \{x + y:\; x,y\in A \} $$ $$ \mathbb{D}(A) = A - A = \{x - y:\; x,y\in A\} $$ Must the sets $\mathbb{S}(A)$ and $\mathbb{D}(A)$ be Borel when $A$ is a Borel set?

I know that it fails when we consider sum of two different sets $A+B$ (see this paper for details).

Choose disjoint perfect sets $P, Q \subseteq [0, 1]$ such that $P \cup Q$ is linearly independent over the field of rationals. So $+ \upharpoonright (P \times Q)$ is a homeomorphism from $P \times Q$ onto a compact subset of $[0, 2]$.

Let $W$ be a Borel subset of $P \times Q$ whose projection on the first coordinate is not Borel. Let $A = \{x + y : (x, y) \in W \}$. Note that $A$ is Borel.

Put $X = A \cup (\{10\} + Q)$ and $Y = A \cup (\{10\} - Q)$. Then both $X, Y$ are Borel as $Q$ is perfect and $A$ is Borel.

Now check that

$$(X - X) \cap (\{10\} - P) = \{10 - x: (\exists y \in Q)((x, y) \in W)\}$$

and

$$(Y + Y) \cap (\{10\} + P) = \{10 + x: (\exists y \in Q)((x, y) \in W)\}$$

It follows that $X - X$ and $Y + Y$ are not Borel.