Practical matter to proving a function is measurable: trouble with definitions... Suppose that I have the simple function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=x$.
To show that $f$ is $(\mathcal{B}_{\mathbb{R}},\mathcal{B}_{\mathbb{R}})$-measurable (where $\mathcal{B}_{\mathbb{R}}$ is the Borel $\sigma$-algebra) I must show that $f^{-1}(E) \in \mathcal{B}_{\mathbb{R}}$ for every $E \subset \mathcal{B}_{\mathbb{R}}$.
I would theoretically have to go through each set $E \in \mathcal{B}_{\mathbb{R}}$ and determine that $f^{-1}(E) \in \mathcal{B}_{\mathbb{R}}$ (I realize I would not actually have to do this, but stay with me...)
Lets say I started with the set $(0,1)$. I know this is a silly question, but how is it possible to show $f^{-1}((0,1)) \in \mathcal{B}_{\mathbb{R}}$ since $f$ is a function and not a correspondence? Even if I went through each singleton $x \in (0,1)$ and showed that $f^{-1}(\{x\}) \in \mathcal{B}_{\mathbb{R}}$, uncountable unions are not necessarily in $\mathcal{B}_{\mathbb{R}}$ so I could not say for sure that $(0,1)$ is in $\mathcal{B}_{\mathbb{R}}$.
Clearly I am having some conceptual difficulties with the idea of measurability. Any help would be appreciated. 
 A: $f$ doesn't map singletons to singletons, it maps reals to reals. The inverse relation $f^{-1}$ induces a map from sets to sets. Sometimes people use square brackets to make the distinction clearer: $f^{-1}[E] = \{x \mid f(x) \in E\}$. Because $f$ is the identity, it's true that the (set map induced by the) inverse maps singletons to singletons – but forget about that! For this $f$, $f^{-1}[E] = \{x \mid f(x) \in E\} = \{x \mid x \in E\} = E$.
A: $f^{-1}((0,1))$ is a set and you have to verify that this set is an element of the set of sets $\mathcal B_{\mathbb R}$.
For example if $f$ is the identity, $f^{-1}((0,1))$ is the open interval $(0,1)$ which belongs to $\mathcal B_{\mathbb R}$.
A: The intersection of preimages (in the domain) is the preimage of the intersection (in the image).
If x belongs to the preimage of the intersection then there is a point y in the intersection on which x is mapped, but then y belongs to all  the sets which intersect in the image, and x belongs to all the preimages (in the domain) if those sets, hence to their intersection. If x belongs to the intersection of the preimages then it belongs to all the sets (in the domain) and its (unique) image y belongs to the image of each set, hence in the intersection and x lies in
the preimage of the intersection.
Similarly for the unions.
It is therefore enough to prove that for any open (or closed) set (or even better: for any collection of the generators of the Borel  sets, such as intervals in the real line) in the image its preimage is a Borel set in the domain. 
