Show that every interval is a Borel set 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? 
 A: 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! 
A: I agree with @Brian M. Scott. You can convince yourself that equality holds by showing the RHS is a subset of the LHS and vice versa. Let us verify that
$$
[a,b]=\bigcap_{n=1}^\infty (a-\tfrac{1}{n},b+\tfrac{1}{n})
$$
is indeed true. 
Therefore, take an $x\in [a,b]$, i.e. $a\leq x\leq b$. Then since $\tfrac{1}{n}>0$ for all $n\in\mathbb{N}$ we have that 
$$
a-\frac{1}{n}<x<b+\frac{1}{n},\quad\text{for all }\,n\in\mathbb N,
$$
and hence $x\in \bigcap\limits_{n=1}^\infty (a-\tfrac{1}{n},b+\tfrac{1}{n})$.
Now, take an $x\in \bigcap\limits_{n=1}^\infty (a-\tfrac{1}{n},b+\tfrac{1}{n})$, i.e.
$$
a-\frac{1}{n}<x<b+\frac{1}{n},\quad\text{for all }\,n\in\mathbb N.
$$
Assume for contradiction that $x\notin [a,b]$. Without loss of generality we can assume that $x>b$. Now, there exist an $n\in\mathbb{N}$ (pick $n$ such that $\tfrac{1}{n}<|x-b|$) such that
$$
x> b+\frac{1}{n}>b
$$
which is a contradiction.
We have shown the two inclusions and conclude that the sets are the same.
