Measure theory makes explicit the distinction between two key concepts: measurability and measure. You may be used to thinking of
$$\int f(x) d\mu(x)$$
as a single object. Two things are needed for this integral to make sense: $f$ needs to be measurable with respect to a $\sigma$-algebra and $\mu$ needs to be a measure defined on that $\sigma$-algebra. Note crucially that measurability of $f$ is not dependent on any kind of measure.
Measurability of a function $f$ is the property that one can distinguish the values that $f$ takes. This is perhaps easiest to understand in the discrete case. Suppose
$$\Omega = \{\omega_1,\omega_2\}, \quad \Sigma = \{\emptyset,\{\omega_1\},\{\omega_2\},\{\omega_1,\omega_2\}\}.$$
What kind of functions $f : \Omega \rightarrow \mathbb{R}$ are measurable with respect to $\Sigma$? Notice that $f$ can take at most $2$ values. If $f(\omega_1) = 1$ and $f(\omega_2) = 3$, then $f^{-1}(A) = \{\omega_1\}$ if $3 \notin A$, $1 \in A$ and $A$ is measurable. However, if $\Sigma = \{\emptyset, \Sigma\}$, even this function $f$ would not be measurable with respect to $\Sigma$. In fact, the only functions which are measurable with respect to the trivial $\sigma$-algebra are the constant ones.
A measure places weights on the sets $E \in \Sigma$. This allows for a notion of approximation. Measures $\mu$ for which $\mu(A) = \inf\{\mu(O) : A \subset O, O \text{ open}\}$ are called outer regular. Measures for which $\mu(A) = \sup\{\mu(K) : K \subset A, K \text{ compact}\}$ are called inner regular. Borel measure on $\mathbb{R}$ is both inner and outer regular. Whether or not the measure is regular depends on the specific $\sigma$-algebra; simply being a $\sigma$-algebra is not sufficient.
To summarise: the coarseness or fineness of the $\sigma$-algebra allows for lesser or greater complexity of functions. There is no notion of approximation of sets in $\Sigma$ at this level; either a set is or is not in $\Sigma$, and $f^{-1}(A) \notin \Sigma$ for some measurable $A$ then $\Sigma$ is not rich enough to "support" $f$.
Measurability of a function $f$ is not related to the measure. The measure places weights on the $\sigma$ algebra. Whether or not it has certain approximation properties depends on the $\sigma$ algebra and the specific measure.
EDIT.
For concreteness, let us pass to the case
$$f : \Omega \rightarrow \mathbb{R},$$
where the image space $\mathbb{R}$ is equipped with the Borel $\sigma$ algebra $\mathcal{B}$ and Lebesgue measure $m$.
For any measurable $A \in \mathcal{B}$, define
$$f^{-1}(A) = \{\omega \in\Omega : f(\omega) \in A\}.$$
To first answer your questions, $f^{-1}([1/2, 3/2]) = \{\omega_1\}$, and $f^{-1}(A) = \emptyset$ if $1 \notin A$ and $3 \notin A$. This may seem strange, but it is exactly what the definition is telling you. What is the set of all $\omega \in \Omega$ such that $f(\omega) \in [1/2, 3/2]$? That set is the singleton $\{\omega_1\}$. This may be confusing because you are looking at a function which takes on finitely many values (2, in this case) on a continuous space. Hopefully what follows will help clarify this issue.
Could you please explain why having a sigma-algebra structure on top of Ω is important in this context?
This is a good question. One reason is that having a $\sigma$-algebra structure allows one to talk about sets of a complexity corresponding to the function in question, on an arbitrary space. To illustrate my point, consider the $f$ given by the picture below.
Let us agree (despite my horrible picture) that $f : [0,4] \rightarrow \mathbb{R}$ is the function given by
$$1\cdot 1_{\color{red}{A}}(x) + 3\cdot 1_{\color{blue}{B}}(x), \quad \color{red}{A} = [0,1) \cup [2,3), \color{blue}{B} = [1,2) \cup [3,4].$$
Is $f$ so much different from the $f$ defined on $\Omega = \{\omega_1,\omega_2\}$? In some sense, yes -- the interval $[0,4]$ contains uncountably many points, while the whole space $\Omega$ only contains two. In another sense, no. The "complexity" of the function has changed little. The $\sigma$-algebra $\sigma(f)$ generated by $f$ is defined to be the smallest $\sigma$-algebra with respect to which $f$ is measurable. In this case, it turns out to be
$$\sigma(f) = \{\emptyset, A, B, [0,4]\}.$$
To see that this is a $\sigma$-algebra, observe that $A^c = B$, $A \cup B = [0,4]$, and $A \cap B = \emptyset$. Thus $A$ and $B$ play the role of $\{\omega_1\}$ and $\{\omega_2\}$ respectively. Is $f$ measurable with respect to this $\sigma$-algebra? Once again, the key is that $f$ takes on finitely many values. Thus $f^{-1}((1,3)) = \emptyset$, while $f^{-1}([1,3]) = [0,4].$ Any other sets can be checked by hand depending on whether or not $1, 3 \in A$.
The thrust of this construction of measure is to make precise what you said:
one classifies functions based on their domains, not on their images!
Indeed. This is the difference between Lebesgue integration and Riemann integration. Whereas to compute
$$\int_0^4 f(x) dx$$
with a Riemann integral, Riemann would have you subdivide many rectangles in the domain needlessly. Instead it is the image space that we should be subdividing. There are only two divisions to be made: $\{\omega \in [0,4] : f(\omega) = 1\}$ and $\{\omega \in [0,4] : f(\omega) = 3\}$, which are $A$ and $B$ respectively. Thus,
\begin{gather*}
\int_0^4 f(x) dx = 1\cdot m(A) + 3\cdot m(B) = 1\cdot 2 + 3\cdot 2 = 8.
\end{gather*}
To come full circle, let us place a measure $\mu$ on $\{\omega_1,\omega_2\}$:
\begin{align*}
\mu(\{\omega_1\}) &= 2 \\
\mu(\{\omega_2\}) &= 2.
\end{align*}
Then, with $\Omega = \{\omega_1,\omega_2\}$ and $f$ defined as in the original post (pre-edit),
$$\int_\Omega f(\omega) d\mu(\omega) = 1\cdot\mu(\omega_1) + 3\cdot\mu(\omega_2) = 8.$$
Observe that the two integrals are handled in essentially the exact same way, because Lebesgue integration and the language of $\sigma$-algebras (i.e., breaking a function $f$ up depending on its image, not its domain) take into account the "complexity" of $f$ and adjust accordingly.
Indeed, the Lebesgue integral is defined by the above procedure. In the case of a function that takes on infinitely many values, such as $f(x) = e^{-x^2}$, we first approximate $f$ in a suitable sense by a function which takes on only finitely many values (called a simple function) and then take limits.
