Showing that half-circle is Borel

Question

Decide if

$$ A= \{(x,y)\in\mathbb{R}^2\mid x^2+y^2\le1, x<y\} $$

is a Borel set.

My guess is that I can use the fact that $\mathscr{B}(\mathbb{R}^n)=\sigma(\mathscr{K})$ where $\mathscr{K}$ is the family of closed sets. I can then somehow show that inclusions $A\subset \sigma(\mathscr{K})$ and $\sigma(\mathscr{K})\subset A$, but I'm not sure if this is the right approach.

Edit:

Using the comments I've rewritten as the set minus of two closed sets

$$ A = \{(x,y)\in\mathbb{R}^2\mid x^2+y^2\le1\}\setminus \{(x,y)\in\mathbb{R}^2\mid x^2+y^2\le1, x\ge y\} $$ I can't see where to go from here.

Answer 1

Notice that $A = \overline{A} \setminus \Delta,$ where $\Delta = \{ (x,y) : x = y \}$ is the diagonal. So, we have written $A$ as the set difference of two closed sets, which are Borel sets. Thus, $A$ is a Borel set.

Answer 2

We observe that $$A=\overline{D(0,1)} \cap \{(x,y) \in \Bbb{R^2}:y-x>0\}$$ where $D(0,1)$ is the unit disk.

Now the set $\{(x,y) \in \Bbb{R^2}:y-x>0\}$ is open because $f^{-1}[(0,+\infty)]$ is open,where $f: \Bbb{R^2} \to \Bbb{R}$ is $f(x,y)=y-x$ which is a continuous function.

So $A$ is the intersection of a closed and an open set so it is a Borel set.