Is the unit circle in $\mathcal{B}(\mathbb{R}^2)$ Is the unit circle in $\mathcal{B}(\mathbb{R}^2)$ ?
The definitions I am working with is that $\mathcal{B}(\mathbb{R}^2)$  is the $\sigma$-Algebra generated by open (equivalently closed, or clopen) rectangles. i.e sets of the form $(a_1,b_1) \times (a_2,b_2)$
Approach 1:
I think maybe I can get there in the limit of unions of lots and lots of rectangles. Here was my approach:
$S =(-1,1) \times (-1,1)$ and then we construct some kind of limit of squares to take away from the square, $S$, that leave us with just the circle?
Approach 2:
Consider $f:(\mathbb{R^2},\mathcal{B}(\mathbb{R}^2)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ by $f(x,y) = x^2 +y^2$
$f$ is clearly continuous and as such is Borel measurable. This means that the pre-image of any measurable set is measurable and hence $f^{-1}(0,1),$ the unit circle, is measurable?
Approach 2 seems to be correct but it feels too simple, are there any mistakes? Would approach 1 provide a fruitful answer?
Any help or even an alternative method of your own would be really helpful!
Happy holidays :)
 A: It is, I will try to make a visual proof, I will use the $\sigma-$alegbra generated by the closed rectangles, wich is the same as you said. Take the disk $B=B(0,1)$ now for $k\in \mathbb N$ imagine all the squares of the form $P_k=\{[n/2^k,(n+1)/2^k]\times[m/2^k,(m+1)/2^k]:m,n\in\mathbb Z\}$ (this is like a grid of squares of length $1/2^k$) and define $A_k$ as the union of all the squares in $P_k$ that are contained in the disk. Since any square in $P_{k}$ can be written as the union of four squares in $P_{k+1}$ you have that $A_k\subset A_{k+1}$, you have that $B=\bigcup A_k$.
To proove this take any point $p$ in $B$, since $B$ is open there is a ball $B_p$ of radius $\varepsilon$ centered in $p$ and contained in $B$, now take $k$ such that $\frac{1}{2^k}<\varepsilon$, then $p$ belongs to a square of $P_{k+1}$ contained in $B_p$:
Take a square $S$ of $P_{k+1}$ containing $p$ (since $P_{k+1}$ covers all $R^2$ the existance of such square is guaranteed) for any point $x$ in $S$ you have that $$d(x,p)\leq \frac{\sqrt{2}}{2^{k+1}}=\frac{1}{2^{k+1/2}}<\frac{1}{2^k}<\varepsilon$$
So $S\subset B_p\subset B$, hence for all $p\in B$ there exist $k$ such that $p\in A_k\subset B$.
