Let $\Sigma$ denote the $\sigma$-algebra generated by the three sets $[0, 1]$, $[2, 4]$, $[3,\infty)$ in $\mathbb R$ and let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra of $\mathbb R$.
I am struggling to understand what the question is asking exactly, any help or recommendations welcome
First, remember that for a map $f \colon (\Omega_1,\mathcal{A}_1) \rightarrow (\Omega_2,\mathcal{A}_2)$ measurable is defined by satisfying $f^{-1}(A) \in \mathcal{A}_1$ for all $A \in \mathcal{A}_2$. In your case it is easy to that every $A \in \Sigma$ is a (finite) union of the sets $$(-\infty,0)\cup (1,2),[0,1],[2,3),[3,4],(4,\infty).$$ We are going to construct a counterexample: Let $E \subset [0,1]$ be a Vitali set. Since $|E| = |\mathbb{R}|$, there is a bijective map $g \colon E \rightarrow [0,1)$. Define $$f(x) = \begin{cases} g(x) & \text{if} \ x \in E \\ 1 & \text{if} \ x \notin E\end{cases}.$$ The preimage of any $A \in \Sigma$ is either empty or already $\mathbb{R}$. Thus $f$ is $\Sigma$-$\Sigma$ measurable. However, $f^{-1}([0,1)) = E \notin \mathcal{B}(\mathbb{R})$, i.e. $f$ is not $\mathcal{B}(\mathbb{R})$-$\mathcal{B}(\mathbb{R})$ measurable.