Borel $\sigma $-algebra of subsets of $[0,1]$ I have been reading Steven Shreve's Stochastic Calculus for finance II. This question is from Chapter $1$ page $4$ General Probability theory. Could someone explain how we got the expression for (a,b) ? Here Formula $1.1.7$ is Uniform (Lebesgue)measure on $[0,1]$ i.e.
$$\mathbb{P} [a,b] = b - a ,\,\, 0 \leq a \leq b \leq 1  $$

Also, what is the process of obtaining Borel $\sigma$ - algebra of subsets of $[0,1]$ ? The text says we start with closed intervals and add everything else required to have a $\sigma$ - algebra? Could anyone give more precise explanation? I know the definition of $\sigma$ - algebra but can't understand how we obtain one beginning from closed intervals ?
 A: If $x \in (a, b)$ then for some $\epsilon > 0$, $a + \epsilon < x < b - \epsilon$. Take $n$ s.t. $\epsilon > \frac{1}{n}$ - then $x \in\left[a + \frac{1}{n}, b - \frac{1}{n}\right]$, so $x \in \bigcup\limits_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right]$ and thus $(a, b) \subseteq \bigcup\limits_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right]$.
In the other direction, if $x \in \left[a + \frac{1}{n}, b - \frac{1}{n}\right]$, then $a < a + \frac{1}{n} \leq x \leq b - \frac{1}{n} < b$ and so $x \in (a, b)$. Thus $\bigcup\limits_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right] \subseteq (a, b)$.
Combining together, we get $(a, b) = \bigcup\limits_{n=1}^\infty \left[a + \frac{1}{n}, b - \frac{1}{n}\right]$.
About obtaining Borel $\sigma$-algebra from closed intervals - formally, it's the smallest (included in any other with such property) $\sigma$-algebra that contains all closed intervals.
Of course, we need to prove that such smallest $\sigma$-algebra exists, but there is standard technique for this.
First, note that $2^{[0, 1]}$ - set of all subsets of $[0, 1]$ is $\sigma$-algebras that contains all closed intervals. Now, let $\mathcal T$ be set of all $\sigma$-algebras on $[0, 1]$ containing all closed interval (it's non-empty,  as $2^{0,1]} \in \mathcal T$).
Intersection of all elements of $\mathcal T$, $\mathcal{B}[0, 1] = \bigcap\limits_{t \in \mathcal T} t$, is again $\sigma$-algebra on $\mathbb{[0, 1]}$ containing all closed intervals (can you prove it?), and it's the smallest such $\sigma$-algebra, because if $t \in \mathcal T$ then $\mathcal{B}[0, 1] \subseteq t$.
It can also be done more constructive with transfinite recursion, but I think non-constructive definition is simpler.
