Show that $f$ is $\mathcal{A}$-measurable if and only if $\{x \in A \mid f(x)=\alpha_i\} \in \mathcal{A}$ for all $i=1,\dots,n$

Question

Let $(X,\mathcal{A})$ be a measure space and $A \subseteq X$. Let $f:A \rightarrow [-\infty,+\infty]$ be a simple function, so that $\{f(x) \mid x \in A\}=\{\alpha_1,\dots,\alpha_n\}$ for some $n \in \mathbb{N}$.
I want to show that $f$ is $\mathcal{A}$-measurable if and only if $\{x \in A \mid f(x)=\alpha_i\} \in \mathcal{A}$ for all $i=1,\dots,n$.
For the first implication, assume $f$ is $\mathcal{A}$-measurable and then write $$\{x \in A \mid f(x) =\alpha_i\}=\{x \in A \mid f(x)\leq \alpha_i\}\cap\{x \in A \mid f(x) \geq\alpha_i\}.$$ These two sets are in $\mathcal{A}$ and so their intersection is in $\mathcal{A}$ as well (since $\mathcal{A}$) is a $\sigma$-algebra. Is this correct?
I am not sure about the other direction.

Answer 1

I do not know exactly which definition of measurablity you are using, but the answer is more or less clear for the other direction: notice that for any Lebesgue measurable set $E\subseteq [-\infty,+\infty]$, we have $E\cap\{\alpha_i\}=(\alpha_{i_k})_{k=1}^{n_E}$, for some indices $i_k$. Therefore, the pre-image $f^{-1} (E)=\cup_{k=1}^{n_E} f^{-1}(\alpha_{i_k})$ is measurable because $f^{-1}(\alpha_{i_k})=\{x\in A:f(x)=\alpha_{i_k}\}$ is measurable for all $k$. The pre-image of all measurable sets are measurable, so $f$ is a measurable function.