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:
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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.
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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?
Best Answer
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 $\mathcal B(\mathbb R^2)$.