prove $\mathscr{g}$ $\subset$ $\mathscr{g}_\delta$, a basic rule given in Measure Theory 2nd Edition written by Donald L. Cohn 1, Let $\mathscr{g}$ be the family of all open subsets of $\mathbb{R}^d$
2, Let $\mathscr{g}_\delta$ be the collection of all intersections of sequences of sets in $\mathscr{g}$
3, And given that each closed subset of $\mathbb{R}^d$ is in $\mathscr{g}_\delta$
Then to prove $\mathscr{g}$ $\subset$ $\mathscr{g}_\delta$, and
$\mathscr{g}_\delta$ $\subset$ $\mathscr{g}_{\delta\sigma}$
This is a rule I think is very basic in measure theory, It is given in the Proposition 1.1.6 of Measure Theory 2nd Edition written by Donald L. Cohn, however the book did not give a detail proof. 
As I think, because of line 3, then $\mathscr{g}_\delta$ contains all closed subsets of $\mathbb{R}^d$, then, if $\mathscr{g}$ $\subset$ $\mathscr{g}_\delta$, means $\mathscr{g}_\delta$ must also contain all open subsets of $\mathbb{R}^d$ because $\mathscr{g}$ contains all open subsets. 
If this is right, how could it be possible that $\mathscr{g}_{\delta\sigma}$ contains some additional subset of $\mathscr{R}^d$ that are not in $\mathscr{g}_\delta$ ?
 A: Obviously $\def\g{\mathscr g}\g \subseteq \g_{\delta} \subseteq \g_{\delta\sigma}$, just take the intersection resp. union of a constant sequence. Moreover, $\g_\delta$ contains all closed sets. Let $A \subseteq \mathbf R^d$ be a closed set, let $U_n :=\{x \in \mathbf R^d : {\rm dist}(A, x) < \frac 1n\}$. Then $U_n \in \g$ and 
$$\bigcap_n U_n = \left\{x \in \mathbf R^d : {\rm dist}(A, x) = 0\right\} = \bar A =  A  $$
Hence $A \in \g_\delta$. So $\g \subsetneq \g_\delta$, as for example $[0,1]^d \in \g_\delta \setminus \g$. Now consider the set 
$$ B := \mathbf Q^d \subseteq \mathbf R^d $$
Then, $B \not\in \g_\delta$, as if $B = \bigcap_n U_n$, for some open $U_n$, the $U_n$ were dense open sets, as $B$ is dense. Therefore, by Baire, $B$ would be of second category, but it isn't, hence $B \not\in \g_\delta$. But $B$ is countable, hence the union of countably many closed sets, namely $B = \bigcup_{x\in B} \{x\}$. Hence $B \in\g_{\delta\sigma}$. Alltogether
$$ \g \subsetneq \g_\delta \subsetneq \g_{\delta\sigma} $$.
