Give an example for which $\sigma (S \cap T) \neq \sigma(S) \cap \sigma(T). $ If $S$ is a collection of subsets of $\mathbb{R}$, we denote by $\sigma(S)$ the smallest $\sigma-$ algebra that contains $S.$
Give an example for which $\sigma (S \cap T) \neq \sigma(S) \cap \sigma(T). $
And I have read the following answer online:
Let $S$ be the collection of Lebesgue measurable sets in $\mathbb{R}.$
Let $B$ be the collection of Borel measurable sets in $\mathbb{R}.$
Let $T = S\setminus B \neq \emptyset$ this is because we have a measurable set , a subset of Borel set, that is not a Borel set.
Then the answer said that, then $S\cap T = \emptyset$(which I donot understand at all) and that $\sigma (S \cap T) = \{\emptyset, \mathbb{R}\}$(which I do not understand why?)And the answer said that also $\sigma(S) \cap \sigma(T) = \{\emptyset, \mathbb{R}\}$(which I do not understand also)
Could anyone explain for me the points I do not understand, it seems to me the example is wrong or lacking details? it is greatly appreciated if I received a correction for this example. 
 A: Let $\mathcal{O}$ denote the set of open sets in $\mathbb{R}$ and $\mathcal{C}$ denote the set of closed sets in $\mathbb{R}$. Then, it is classical that $\sigma(\mathcal{O})=\sigma(\mathcal{C})=\mathcal{B}(\mathbb{R}),$ with the latter denoting the Borel $\sigma$-algebra. Accordingly $\sigma(\mathcal{C})\cap \sigma(\mathcal{O})=\mathcal{B}(\mathbb{R})$.
However, $\mathcal{O}\cap \mathcal{C}=\{\emptyset,\mathbb{R}\},$ since $\mathbb{R}$ is connected, so $\mathcal{O}\cap\mathcal{C}$ already is a $\sigma$-algebra, and hence $\sigma(\mathcal{O}\cap \mathcal{C})=\{\emptyset,\mathbb{R}\}$. This clearly isn't the Borel $\sigma$-algebra.
A: Simpler (and more correct) example: Let $S$ be the set of all open  intervals with rational endpoints and $T$ the set of open intervals with irrational endpoints. Then clearly $S\cap T=\emptyset$ and so $\sigma(S\cap T)=\{\emptyset,\Bbb R\}$ (which is the smallest $\sigma$-algebra at all). On the other hand, $\sigma(S)\cap\sigma(T)$ is much larger as for example $$(\sqrt 2,\sqrt3)\in T\subseteq \sigma(T)$$ and also $$(\sqrt 2,\sqrt 3)=\bigcup_{\sqrt 2<a<b<\sqrt 3\atop a,b\in\Bbb Q }(a,b)\in \sigma(S).$$
(Bonus: Do you see why in fact $\sigma(S)=\sigma(T)$ and which well-known algebra that is?)
