This is a question from an old exam. I'd like to see if my answer is correct. I'd appreciate any suggestion. Thanks :)
Let $f: (\mathbb{R},\mathscr{S}) \to (\mathbb{R}, \mathscr{B}(\mathbb{R}))$ defined as follows
$$f(x):=3\cdot \mathbb{1}_{[1,3]}(x)+2\cdot \mathbb{1}_{[2,4]}(x)$$
where $\mathscr{B}(\mathbb{R})$ is the Borel $\sigma$-algebra of the real numbers, and $\mathbb{1}_A(x)$ is the indicator function, i.e., if $x$ belongs to $A$, $\mathbb{1}_A(x)=1$ and it is zero otherwise.
What is the minimum $\sigma$-algebra $\mathscr{S}$ that makes $f$ a measurable function (write explicitly the sigma algebra)?
Answer: Let $B_1=[1,3]$ and $B_2=[2,4]$. Now let $\mathscr{D}$ be the collection of all the intersection $C_1\cap C_2$, where either $C_i=B_i$ or $C_i=B_i^c$ for $i\in \{1,2\}$. Then
$$\mathscr{D}=\{\,[1,2),\,[2,3],\,(3,4],\,(-\infty,1)\cup (4,+\infty)\,\}$$
Now, we enumerate the elements in $\mathscr{D}$, i.e., $\mathscr{D}=\{D_j: \text{ for } j= 1,\ldots, 4 \}$, and define $\mathscr{A}$ as the collection of all the possible unions of the elements in $\mathscr{D}$, that is $$A\in \mathscr{A} \iff A=\bigcup\bigg\{D_j : j\in J \text{ where } J \subset\{1,2,3,4\}\bigg\}$$
Thus, we have the follwing \begin{align*}\mathscr{A}= \bigg \{ &\varnothing,\,[1,2),\,[2,3],\,(3,4],\,(-\infty,1)\cup (4,+\infty),\,[1,3],\, [1,2)\cup (3,4],\,(-\infty,2)\cup (4,+\infty),\\ &[2,4],\,(-\infty,1)\cup [2,3]\cup (4,+\infty),\,(-\infty,1)\cup (3,+\infty), [1,4],\,(-\infty,3]\cup (4,+\infty),\\ & (-\infty,2)\cup (3,+\infty),\,(-\infty,1)\cup [2,+\infty),\, \mathbb{R} \bigg\} \end{align*}
Now clearly $\mathscr{A}$ is a $\sigma$-algebra (it is closed under countable unions, complementation and also $\mathbb{R}$ belongs to it). On the other hand, the function $f$ can be written by elements in $\mathscr{D}$ (which are a partition of the reals) in the following way
$$f(x)=3 \cdot \mathbb{1}_{[1,2)}(x)+5\cdot \mathbb{1}_{[2,3]}(x)+2 \cdot \mathbb{1}_{(3,4]} (x)$$
Therefore, in the above expression of $f$ it is clear that $\mathscr{A}$ makes $f$ measurable.The minimum sigma algebra $\mathscr{S}$ that makes $f$ measurable, must contain $[1,2),\,[2,3],\,(3,4]$. Then $\mathscr{D}$ belongs to $\mathscr{S}$, since $(-\infty,1)\cup (4,+\infty)=[1,4]^c$, and so $\mathscr{A}\subset \mathscr{S}$. The other inclusion is easy by minimality of $\mathscr{S}$. Then $\mathscr{S} =\mathscr{A}$ as desired.
Yes, I think your proof looks fine.
In general, when you have a $\sigma$-algebra generated by a finite number of sets, it will always produce a partition of your space ($\mathbb{R}$ here) into a finite number $n$ of subsets, and the $\sigma$-algebra will correspond to all unions of those subsets and contain a total of $2^n$ sets.