TITLE: Show that every interval is a Borel set
QUESTION [12 upvotes]: Show that every interval is a Borel set.
My textbook states:

The intersection of all the $\sigma$-algebras of subsets of $\mathbb{R}$ that contain the open sets is a $\sigma$-algebra called the Borel $\sigma$-algebra; members of this collection are called Borel sets. 

My answer:
Since $(-\infty, a)$, $(a,b)$, $(a,\infty)$ are open, they must be Borel. For any interval of the following forms, we also see that they can be represented by open intervals.
$$
\begin{align}
[a,b]&=\bigcap^{\infty}_{n=1} (a-1/n, b+1/n)\\
(a,b]&=\bigcap^{\infty}_{n=1} (a, b+1/n)\\
[a,b)&=\bigcap^{\infty}_{n=1} (a-1/n, b)
\end{align}
$$
Do you think my answer is correct?

REPLY [2 votes]: You could have tried it easier. 
For example, 
Finte sets are are closed in R . The compliments of which are union of open sets, are borel and hence any finite set is borel. You can write a clopen or a closed interval as union of an open interval and a finite set and so they are borel!