Question on Borel Measurable Functions and Borel Sigma Algebras. On an assignment, we have been given a the following setup:
Let $ f:A\rightarrow \bar{\mathbb{R}}$ be a borel measurable function. prove that if $f$ is Borel measurable and $B$ is a Borel set, then $f^{-1}(B)$ is a Borel set.
The definition of Borel Measurability we were given is as follows:
"A function $f : \mathbb{R} \to \mathbb{R}$ is said to be Borel measurable provided
its domain A ⊆ R is a Borel set and for each c, the set {$x ∈ A : f (x) < c$} is a Borel
set. 
We were not given any description of where this set lives, I'm assuming $B \subset \mathbb{R}$ but this may be incorrect. I figure we need to show that the set {${B \subset \mathbb{R}:f^{-1}(B)}$ is a Borel set} is a sigma algebra but I am unsure how to do this. I know its a little silly because all you need to do is check that the definitions of a sigma algebra are met but showing those things is proving more difficult than i anticipated. We may also use the fact that borel measurable functions are Lebesgue measureable. Any help would be greatly appreciated!
 A: Note that for $b \in \Bbb{R}$
$f^{-1}((-\infty,b])=A \setminus f^{-1}(b,+\infty)$ which is a Borel set.
The family of sets of the form $(-\infty,b]$ with the empty set,generate the Borel sigma algebra.
So if you show that $A=\{B \subseteq R: f^{-1}(B) \text{is Borel}\}$ is a sigma algebra then you are done since this sigma algebra contains the sets of the form $(-\infty,b]$ and the empty set,so it will contain the Borel sigma algebra by definition.
To show that $A$ is sigma algebra,just use the fact that the inverse image of a union is the union of the inverse images and 
if $B \subseteq \Bbb{R}$ then  $f^{-1}(B^c)=A \setminus f^{-1}(B)$
A: The Borel $\sigma$-algebra of some topological space $A$ is defined as the smallest $\sigma$-algebra that contains all the open sets of $A$.
When we says that a function $f:A\to\overline{\Bbb R}$ is Borel measurable we are assuming the standard topology in $\overline{\Bbb R}$ and that if $C\subset \overline{\Bbb R }$ is a Borel set then $f^{-1}(C)$ is Borel in the induced Borel $\sigma$-algebra of $A$.
By the properties of $f^{-1}$ respect to set operations it can be shown that it is enough to say that $f^{-1}(C)$ is Borel in $A$ for any $C$ of the form $[-\infty,a)$, as is generally explained in any analysis textbook that cover an introduction to Lebesgue integration theory.
