1.2 Proposition - $B_{\mathbb{R}}$ is generated by each of the following:
a.) the open intervals: $\epsilon_1 = \{(a,b): a<b\}$
b.) the closed intervals: $\epsilon_2 = \{[a,b]: a<b\}$
c.) the half-open intervals: $\epsilon_3 = \{(a,b]: a < b\}$ or $\epsilon_4 = \{[a,b): a < b\}$
d.) the open rays: $\epsilon_5 = \{(a,\infty): a\in \mathbb{R}\}$ or $\epsilon_6 = \{(-\infty,a): a\in \mathbb{R}\}$
e.) the closed rays: $\epsilon_7 = \{[a,\infty): a\in \mathbb{R}\}$ or $\epsilon_8 = \{(-\infty,a]: a\in \mathbb{R}\}$
You have prove a,b, and c. So we need only to prove d. and e.
Proof of d:
Clearly for any $A \in \epsilon_5$, $A$ is an open set and so, we have that $A\in B_{\mathbb{R}}$. So we have proved that $\epsilon_5 \subset B_{\mathbb{R}}$, and so, being $\sigma(\epsilon_5)$ the $\sigma$-algebra generated by $\epsilon_5$, we have
$$\sigma(\epsilon_5) \subset B_{\mathbb{R}}$$
Now note that for any $b\in \mathbb{R}$, we have
$$[b, \infty ) = \bigcap_{n=1}^\infty (b-(1/n), \infty)\in \sigma(\epsilon_5)$$
So, for any $a, b\in \mathbb{R}$, $a<b$, we have
$$ (a,b) = (a, \infty) \setminus [b, \infty ) \in \sigma(\epsilon_5)$$
Since all open set in \mathbb{R} can be expressed as a countable union of intervals $(a,b)$, where $a, b\in \mathbb{R}$, then, being $\tau$ the set of open set of $ \mathbb{R}$, we have that
$$ \tau \subset \sigma(\epsilon_5)$$
So we have
$$B_{\mathbb{R}} = \sigma(\tau) \subset \sigma(\epsilon_5)$$
So we can conclude that
$$ \sigma(\epsilon_5) = B_{\mathbb{R}}$$
The proof for $\epsilon_6$ is totally similar.
Clearly for any $A \in \epsilon_6$, $A$ is an open set and so, we have that $A\in B_{\mathbb{R}}$. So we have proved that $\epsilon_6 \subset B_{\mathbb{R}}$, and so, being $\sigma(\epsilon_6)$ the $\sigma$-algebra generated by $\epsilon_6$, we have
$$\sigma(\epsilon_6) \subset B_{\mathbb{R}}$$
Now note that for any $a\in \mathbb{R}$, we have
$$(-\infty , a] = \bigcap_{n=1}^\infty (-\infty , a+(1/n)]\in \sigma(\epsilon_6)$$
So, for any $a, b\in \mathbb{R}$, $a<b$, we have
$$ (a,b) = (-\infty, b) \setminus (-\infty , a] \in \sigma(\epsilon_6)$$
Since all open set in \mathbb{R} can be expressed as a countable union of intervals $(a,b)$, where $a, b\in \mathbb{R}$, then, being $\tau$ the set of open set of $ \mathbb{R}$, we have that
$$ \tau \subset \sigma(\epsilon_6)$$
So we have
$$B_{\mathbb{R}} = \sigma(\tau) \subset \sigma(\epsilon_6)$$
So we can conclude that
$$ \sigma(\epsilon_6) = B_{\mathbb{R}}$$
Proof of e:
Clearly for any $A \in \epsilon_7$, $A$ is a closed set and so, we have that $A\in B_{\mathbb{R}}$. So we have proved that $\epsilon_7 \subset B_{\mathbb{R}}$, and so, being $\sigma(\epsilon_7)$ the $\sigma$-algebra generated by $\epsilon_7$, we have
$$\sigma(\epsilon_7) \subset B_{\mathbb{R}}$$
Now note that for any $a\in \mathbb{R}$, we have
$$(a, \infty ) = \bigcup_{n=1}^\infty [a+(1/n), \infty)\in \sigma(\epsilon_7)$$
So, for any $a, b\in \mathbb{R}$, $a<b$, we have
$$ (a,b) = (a, \infty) \setminus [b, \infty ) \in \sigma(\epsilon_7)$$
Since all open set in \mathbb{R} can be expressed as a countable union of intervals $(a,b)$, where $a, b\in \mathbb{R}$, then, being $\tau$ the set of open set of $ \mathbb{R}$, we have that
$$ \tau \subset \sigma(\epsilon_7)$$
So we have
$$B_{\mathbb{R}} = \sigma(\tau) \subset \sigma(\epsilon_7)$$
So we can conclude that
$$ \sigma(\epsilon_7) = B_{\mathbb{R}}$$
The proof for $\epsilon_8$ is totally similar.
Clearly for any $A \in \epsilon_8$, $A$ is a closed set and so, we have that $A\in B_{\mathbb{R}}$. So we have proved that $\epsilon_8 \subset B_{\mathbb{R}}$, and so, being $\sigma(\epsilon_8)$ the $\sigma$-algebra generated by $\epsilon_8$, we have
$$\sigma(\epsilon_8) \subset B_{\mathbb{R}}$$
Now note that for any $b\in \mathbb{R}$, we have
$$(-\infty, b ) = \bigcup_{n=1}^\infty (-\infty, b-(1/n)]\in \sigma(\epsilon_8)$$
So, for any $a, b\in \mathbb{R}$, $a<b$, we have
$$ (a,b) = (-\infty, b) \setminus (- \infty, a] \in \sigma(\epsilon_8)$$
Since all open set in \mathbb{R} can be expressed as a countable union of intervals $(a,b)$, where $a, b\in \mathbb{R}$, then, being $\tau$ the set of open set of $ \mathbb{R}$, we have that
$$ \tau \subset \sigma(\epsilon_8)$$
So we have
$$B_{\mathbb{R}} = \sigma(\tau) \subset \sigma(\epsilon_8)$$
So we can conclude that
$$ \sigma(\epsilon_8) = B_{\mathbb{R}}$$
Best Answer
Yes, that is a correct solution. However, I recommend writing it up in the other direction and in an assertive manner. Maybe like so:
Since for all $a < b$ we have $(a,b] = (a,+\infty) \cap (b,+\infty)^c \in \sigma(\mathcal{H})$. It follows that $\mathcal{I} \subset \sigma(\mathcal{H})$, and hence $\sigma(\mathcal{I}) \subset \sigma(\mathcal{H})$. By [cite lecture] we know that $\mathcal{B} = \sigma(\mathcal{I})$, thus $\mathcal{B}\subset \sigma(\mathcal{H}) \subset \mathcal{B}$, q.e.d.