difference between $f^{-1}([-\infty ,a))$ and $f^{-1}(-\infty ,a)$? In the book "Real-Analysis" of Stein and Shakarshi, they say that $f$ is measurable if $f^{-1}([-\infty ,\alpha ))$ is measurable for all $\alpha \in\mathbb R$ but in my course a function $f$ is said to be measurable if $f^{-1}((-\infty ,\alpha ))$ is measurable for all $\alpha \in\mathbb R$. Are both definitions equivalent? And if yes, why?
Furthermore, what sets are open in $[-\infty ,\infty ]$? Are they the sets of the form $]a,b[$, $[-\infty ,a)$ and $(b,+\infty ]$? And in $[-\infty ,\infty ]$ is $(-\infty ,a)$ still open ? 
 A: Sometimes, particularly in measure theory, people choose to consider the extended real line, or the real line with two points called $-\infty$ and $\infty$ adjoined. This is often written $[-\infty, \infty]$. This is a useful space to study since positive measures are functions from a $\sigma$-algebra to $[0,\infty]$, and integration by positive measures can yield $-\infty$ or $\infty$.
The extended real line is typically ordered by putting $\infty$ as the largest element and $-\infty$ as the smallest element. Once you have this order on the $[-\infty, \infty]$, the order topology can be put on the space. The open and closed sets are then the standard ones with respect to the order topology. It's a nice topology. The subspace topology of $(-\infty, \infty)$ with respect to it is the standard topology on $\mathbb{R}$, and $[-\infty, \infty]$ is compact and Hausdorff. In fact, it's homeomorphic to $[0,1]$.
However, $[-\infty, \infty]$ loses much of the algebraic niceness of $\mathbb{R}$. The expression $- \infty + \infty$ cannot be coherently defined. For this reason, attention is often restricted to functions on $[0,\infty]$ (like positive measures), where addition and multiplication can be nicely defined (put $0 * \infty = 0$).
In the context of your book, they are defining measurability of extended real-valued functions; in class, they are defining it for real-valued functions. Since the subspace topology of $(-\infty, \infty)$ in $[-\infty, \infty]$ is the standard topology on $\mathbb{R}$, these definitions coincide for all real-valued functions. Your book is just being a bit more general.
A: In this context it might be useful to formulate a more general statement that catches both.
Let it be that $(X,\mathcal M)$ is a measurable space and that $(Y,\tau)$ is a topological space. So here $\mathcal M$ denotes a $\sigma$-algebra of subsets of $X$ and $\tau$ denotes a topology of subsets of $Y$. Then on $Y$ we have the Borel $\sigma$ algebra generated by $\tau$ and is denoted by $\sigma(\tau)$. 
Now let it be that $f:X\to Y$ is a function and $\mathcal A$ is a collection of subsets of $Y$.
Then the following is true:$$\text{If }\tau\subseteq\sigma(\mathcal A)\text{ then: }f:(X,\mathcal M)\to(Y,\sigma(\tau))\text{ is measurable if and only if }f^{-1}(\mathcal A)\subseteq\mathcal M\tag1$$
If we are dealing with a second countable topology and $\mathcal V$ denotes a subbase for this topology then $(1)$ can  be replaced by:$$\text{If }\mathcal V\subseteq\sigma(\mathcal A)\text{ then: }f:(X,\mathcal M)\to(Y,\sigma(\tau))\text{ is measurable if and only if }f^{-1}(\mathcal A)\subseteq\mathcal M\tag2$$
This follows from $(1)$ combined with the fact that $\tau\subseteq\sigma(\mathcal V)$ in this situation.
The expression $f^{-1}((-\infty,a))$ in your question indicates that we are dealing with special case $Y=\mathbb R$ equipped with usual (order) topology and $\mathcal A=\{(-\infty,a)\mid a\in\mathbb R\}$. The topology is second countable and as a subbasis serves the collection of sets that can be written as $(-\infty,a)$ or $(a,\infty)$. 
The expression $f^{-1}([-\infty,a))$  indicates that we are dealing with special case $Y=[-\infty,\infty]$ and $\mathcal A=\{[-\infty,a)\mid a\in\mathbb R\}$. Also here $Y$ is equipped with the so-called order topology and as a subbasis serves the collection of sets that can be written as $[-\infty,a)$ or $(a,\infty]$.
In the first case $(2)$ can be used to prove that a function $f:X\to \mathbb R$ that satisfies $f^{-1}((-\infty,a))\in\mathcal M$ for every $a\in\mathbb R$ is measurable. Actually it is enough already to prove that also $f^{-1}((a,-\infty))\in\mathcal M$ for every $a\in\mathbb R$.
Same story for second case.

Note:
This answer does not provide a full proof (yet) of $(1)$ and $(2)$.
