Show that $A=\{(x,y) \in \mathbb R^2 \mid x^2+y^2 \leq 1 , \, x<y\}$ is Borel

Question

In my homework I want to show that $A=\{(x,y) \in \mathbb R^2\mid x^2+y^2 \leq 1 , \, x<y\}$ is Borel and determine if A is open or closed

My idea was to say that:

$A$ is the intersection or all $\{(x,y) \in \mathbb R^2\mid x^2+y^2 \leq 1 $ but I don't know how to handle the x

Also it seems closed on the half-circle but open on the $x<y$ but it doesn't make sense that it is both open and closed.

Any help would be appreciated

Answer 1

$A_1=\{(x,y)\in\mathbb{R}^2:x^2+y^2\le 1\}$ is a Borel set ($\because$ it is closed), $A_2=\{(x,y)\in\mathbb{R}^2:x-y<0\}$ is a Borel set ($\because$ it is open). Then $A=A_1\cap A_2$ is also a Borel set.