Borel $σ$-algebra generated by intervals on real line Show that Borel $σ$-algebra $B_{\mathbb R} = \sigma(A)$, when $A$ is:
(a) $\{(a, b) : a, b \in \mathbb{R} \}$, 
(b) $\{(-\infty, a) : a \in  \mathbb{R}\}$, 
(b) $\{(-\infty, a] : a \in \mathbb{R}\}$,
(d) $\{(a,\infty) : a \in \mathbb{R}\}$.
The proof do not have to be strict.
My solution:
For (a) I found out that the interval $(a,b)$ can be written like:
$(a,b)=\bigcup_{n=1}^\infty (a, b-\frac{1}{n})$ or $(a,b)=\bigcap_{n=1}^\infty (a, b-\frac{1}{n}).$
But, however, I have no idea how to show that $B_{\mathbb R} = \sigma\left(\{(a, b) : a, b \in \mathbb{R}\}\right)$.
 A: Let the sets $A_a$ to $A_d$ be the collections of intervals used in part (a) to (d) resp. 
Let $\mathcal{S}$ be a $\sigma$-algebra on $\mathbb{R}$ that contains $A_a$. Then it contains $A_b$, as for all $a$:
$$(-\infty, a) = \bigcup_{n=1}^\infty (a-n,a) \in \mathcal{S}$$ 
If $\mathcal{S}$ contains $A_b$ it contains all of $A_c$ as, for all $a$: 
$$(-\infty, a] = \bigcap_{n=1}^\infty (-\infty, a+\frac1n) \in \mathcal{S}$$ 
If $\mathcal{S}$ contains $A_c$ it contains $A_d$ too, as for all $a$:
$$(a,+\infty)= (-\infty,a]^\complement \in \mathcal{S}$$
If $\mathcal{S}$ contains $A_d$, then it contains $A_a$ in a few steps, to prove that $(a,b)$ in it: 
$$(a,c] = (a, +\infty) \cap (c, +\infty)^\complement \in \mathcal{S}$$ 
for all $c > a$, and then 
$$(a,b) = \bigcup_{n=k}^{\infty} (a,b-\frac1n] \in \mathcal{S}$$
for $k$ large enough.
So if a $\sigma$-algebra $\mathcal{S}$ contains one of the families $A_a$ to $A_d$, it contains all of them and so $\mathcal{B}_\mathbb{R}$ is contained in all of them, as it is the smallest $\sigma$-algebra that contains all of these intervals. 
Also by definition $\sigma(A_p) \subseteq \mathcal{B}_\mathbb{R}$ as the latter is one $\sigma$-algebra that contains $A_p$ ($p = a,b,c,d$) and $\sigma(A_p)$ is the smallest such.
So $\sigma(A_a) = \sigma(A_b) = \sigma(A_c) = \sigma(A_d) = \mathcal{B}_\mathbb{R}$.
A: For each $A$ it is clear that the $\sigma$-algebra generated by $A$ is contained in the Borel $\sigma$-algebra.
For the other inclusion, you have to show that a $\sigma$-algebra containing $A$ must contain all intervals.
E.g. in (1), you have $[a,b]=\bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})$ hence all closed intervals are in $\sigma(A)$. All the open intervals are there by definition. For half-open intervals, you can use a similar argument or look at $[a,b)=(a-1,b)\cap[a,b]$ etc.
In (2), it is enough to show that all finite open intervals are in $\sigma(A)$, and then you can use (1). First you see that $[a,b)=(-\infty,b)\cap(-\infty,a)^{c}\in \sigma(A)$, and now $(a,b)=\bigcup[a+\frac{1}{n},b)$.
