I'm working through the exercises of Durrett, and I'm confused about the second part of this particular exercise.
We first show $\sigma(\mathcal{S}_d) \subseteq \mathcal{R}^d $ by showing that $\mathcal{S}_d \subseteq \mathcal{R}^d$. Let $A = (a_1, b_1] \times \cdots \times (a_d, b_d] \in \mathcal{S}_d$.
I've found a detailed solution here, (the different numbering of the exercise is due to different edition of the book), and I understand the first part of the exercise, but confused about the second part.
Let $A$ be an open set in $\mathcal{R}^d$. Let $A_{n, i} \in S_n$ be a sequence of all sets in $S_n$ that are contained in $A$. Let $B = \cup_{n,i} A_{n, i}$ then by definition of $A_{n, i}$, we have $\cup_{n, i} A_{n, i} \subseteq A$. We claim that all points in $A$ are contained in some $A_{n, i}$ to establish the reverse containment. Suppose $a \in A$. Since $A$ is an open set, there is a cube centered at $a$ that is contained in $A$. It is clear that there is an $n$ for which there exists a set $A_{n, i}$ that is contained in $A$. Therefore, $\cup_{n, i} A_{n, i} = A$. Since the left hand side is a countable union of sets in $\mathcal{S}_d$, $A \in \sigma(\mathcal{S}_d)$.
The introduction of such a sequence is very confusing for me, I understand what is n, but as we are talking about $ \mathbb{R}^d $, I'm confused about the need for the iterator i, and what it is exactly coding. I understand that n encodes for the granularity, but $i$ doesn't add any information on what the coefficients will be, at least for me.
I tried to step through that, but I found the entire explanation very convoluted, so I would appreciate an alternative explanation, and I would be also be interested in the background intuition for this problem.
I finally understood your problem... (but be rigorous in your questions, because you neither defined $\mathcal S_d$ nor $\mathcal R^d$, and we are not supposed to guess them). Instead of doing the proof in $\mathbb R^n$, do it in $\mathbb R$ (the principle is exactly the same and easier to see).
As you understood (so I won't prove it), $\mathcal S\subset \mathcal R$, and thus $\sigma (\mathcal S)\subset \mathcal R$.
To show the converse, we want to show that open sets are in $\sigma (\mathcal S)$. Indeed if $\mathcal O=\{\text{open sets of }\mathbb R\}$ and $\mathcal O\subset \sigma (\mathcal S)$, then $$\mathcal R=\sigma (\mathcal O)\subset \sigma (\mathcal S),$$ what will prove the claim.
So, let $A$ be an open set of $\mathbb R$. What you want is to write $A$ as a countable union of elements of $\mathcal S$. So, let $$\mathcal U=\{(a,b]\mid (a,b]\subset A\text{ and } a,b\in \mathbb Q \}.$$ Notice that $\mathcal U\subset \mathcal S$ and that $\mathcal U$ is countable. Moreover, $$\bigcup_{U\in \mathcal U}U\subset A.$$
For the converse inclusion, let $a\in A$. Since $A$ is open, there exist $\delta >0$ s.t. $(a-\delta ,a+\delta )\subset A$. By density of $\mathbb Q$ in $\mathbb R$, we can find $c,d\in\mathbb Q$ s.t. $$a-\delta <c<a<d<a+\delta.$$ Therefore, $$a\in \underbrace{(c,d]}_{\in \mathcal U}\subset A,$$ i.e. $\displaystyle a\in \bigcup_{U\in \mathcal U}U$. Therefore, $\displaystyle A=\bigcup_{U\in \mathcal U}U$ and thus $A\in \sigma (\mathcal S) $ as wished.