How to understand that $\mathcal F$ is a $\sigma$-field containing all intervals? Since I cannot understand the definition of Borel $\sigma$-field, which is defined as $$\mathcal B=\bigcap\{\mathcal F:\mathcal F \text{ is a }\sigma\text{ -field containing all intevals}\}.$$
We say that $\mathcal B(\mathbb{R} )$ is the smallest $\sigma$-field generated by all intervals and we call the elements of $\mathcal B$ Borel sets.
        $\mathbb R$ is a set containing all real numbers. $\mathcal F$ is the $\sigma$-field on $\mathbb R$, then $\mathcal F \text{ is a proper }\text{collection of subsets of }\mathbb R$ satisfying 3 conditions of the definition of the $\sigma$-field. But how can we say that $\mathcal F$ contains all the intervals? Some examples? 
Thank you!
 A: Let $\mathcal A$ be the collection of $\sigma$-fields that contain the (open, closed, half-open/closed) intervals. That is, if $\mathcal F \in \mathcal A$, then $\mathcal F$ is a $\sigma$-field, and for every $a,b\in\mathbb R$, we have that $(a,b),[a,b],[a,b),(a,b] \in \mathcal F$. We then define the Borel $\sigma$-field to be $\mathcal B = \cap \mathcal A$. (Note: We could have taken $\mathcal A$ to just be a collection of $\sigma$-fields that contain only the open intervals, or even just the closed intervals, etc. This makes no difference in defining the Borel $\sigma$-field.(*))
Fix some $\mathcal F \in \mathcal A$. Your first question asks, 

How can we say that $\mathcal F$ contains all the intervals?

I'm not sure this question is very meaningful because the answer is just that we can say that $\mathcal F$ contains all the intervals because that's how we've defined $\mathcal F$.
To illustrate why I think the question isn't very meaningful, consider the following exercise. Let $E$ be the set of all real numbers greater than $7$. You could just as well ask how can we say that $E$ contains all the real numbers greater than $7$. The answer, in this case, is the same. We can say that $E$ contains all the real numbers greater than $7$ because we've defined $E$ to have this property. (**)
For your second question, one example of a $\sigma$-field that contains all the intervals, as I pointed out in a comment is the power set of $\mathbb R$.
(*) This is because if $\mathcal F$ is a $\sigma$-field that contains all the open intervals, then it contains all the other intervals as well, because they can be written as a countable union/intersection of sets already in $\mathcal F$. For example, we have that $$ [0,1] = \bigcap_{n=1}^{\infty} \left(0-\frac 1n , 1 + \frac 1n \right) \in\mathcal F. $$
(**) Of course, you may at this point ask what makes this way of defining sets valid, and that would be an excellent question to ask. This then gets into issues dealt with in the field of set theory, which you can look into if you're interested. However, for most of mathematics outside of set theory, we typically take for granted that we can define sets in this way without worrying too much about how exactly this procedure works.
