Borel sigma algebra on $\mathbb{R}^2$. Consider the family of all circles on plane ($S = \{S_{x_0, y_0, c} : (x-x_0)^2 + (y-y_0)^2 = c^2)\}$). We want to know that $\sigma(S) \ne \mathcal{B}(\mathbb{R}^2)$.
Intuitively I have two explanations:

*

*It looks like if it was true, then the family $\{x:y = c\}\cap S$ should generate $\mathcal{B}(\mathbb{R})$. But at the same time, these intersections are just singletons (or it's collections), hence we can't generate Borel sigma-algebra.


*The second idea is that there should be an analogy of countable-countable sigma-algebra on plane, which generated by $S$ and doesn't coincides with $\mathcal{B}(\mathbb{R}^2)$.
Any hint or directions to think about?
 A: There is a likely a more elementary method, but here's a quick proof using existence of the Lebesgue measure:
Let $\mu$ denote the Lebesgue measure on $\mathbb R^2$. Then $\mu(E) = 0$ for all $E \in S$. Consider
$$ \mathcal E = \{E \in \mathcal B(\mathbb R^2) : \text{$\mu(E) = 0$ or $\mu(E^c) = 0$} \}. $$
It's straightforward to show $\mathcal E$ is a $\sigma$-algebra. Further $S \subset \mathcal E$ thus $\sigma(S) \subset \mathcal E$. However, $\mathcal E$ is a strict subset of $\mathcal B(\mathbb R^2)$. For example $[0,1] \times [0,1]$ is in $\mathcal B(\mathbb R^2)$ but not in $\mathcal E$. Thus $\sigma(S)$ is also a strict subset of $\mathcal B(\mathbb R^2)$.
A: Let me prove a result which includes your question as a special case. In the below, $\lambda$ denotes $n$-dimensional Lebesgue measure on $\mathbb{R}^n$.
Proposition. Let $\mathcal{S}$ be a collection of subsets of $\mathbb{R}^n$ such that $\lambda(S)=0$ for all $S\in\mathcal{S}$. Then $\sigma(\mathcal{S})\neq \mathcal{B}(\mathbb{R}^n)$.
Proof. Let $\mu$ be a "nice enough" probability measure on $\mathcal{B}(\mathbb{R}^n)$ such that $\mu\ll\lambda$ (i.e. $\mu$ is absolutely continuous with respect to $\lambda$). For a concrete example, we can take $\mu$ to be
$$\mu(A)=\int_{A}\prod_{i=1}^{n}\frac{1}{\sqrt{2\pi}}e^{-x_i^2/2}d\lambda,\enspace A\in\mathcal{B}(\mathbb{R}^n).$$
Let $\mathcal{T}:=\{A\in \mathcal{B}(\mathbb{R}^n):\mu(A)=0\text{ or }\mu(A)=1\}$, and note that $\mathcal{T}$ is a sub-$\sigma$-algebra of $\mathcal{B}(\mathbb{R}^n)$. Since $\mathcal{S}\subseteq \mathcal{T}$, we have $\sigma(\mathcal{S})\subseteq\mathcal{T}$. Also, it is easy to see that $\mathcal{T}$ is strictly contained in $\mathcal{B}(\mathbb{R}^n)$ (for example $A=[0,\infty)^n$ has $\mu(A)=2^{-n}$). So $\sigma(\mathcal{S})$ cannot be all of $\mathcal{B}(\mathbb{R}^n)$.
To deduce the answer to your question, we just have to take $\mathcal{S}$ to be the collection of circles in $\mathbb{R}^2$ and apply the proposition above.
