Let $(E,\cal{E})$ be a measurable space, and let $f:E\rightarrow\mathbb{R}$ be an elementary function, i.e. $f=\sum_{n=1}^{\infty}a_n\mathbb{I}_{A_n}$ where $\{A_n\}_{n=1}^{\infty}$ is a measurable partition of $E$, and, for each $n\geq1$, $ \,\mathbb{I}_{A_n}$ is the indicator function of $A_n$.
Prove that $f$ is $\mathcal{E}$-measurable, i.e. $f^{-1}(-\infty,a)\in\mathcal{E} \;$ $\forall{a} \in \mathbb{R}$.
(This is taken from exercise 2.26 of Chapter 1 of "Probability and Stochastics" by Erhan Çınlar.)
My attempt
Fix $a\in\mathbb{R}$ and consider any $b \in (-\infty, a) \;$ such that $f(x)=b \;$ for some $x\in E \;$.
By definition of $f$, there are at most countably many $A_{n,b} \in \mathcal{E}$ such that $f(x)=b \;$ for some $x \in A_{n,b}$. So, there is $A_b=\bigcup_{k=1}^{\infty}A_{n_k,b} \;$ such that $f^{-1}\{b\}=A_b$. And $A_b$ is measurable as it is a countable union of measurable sets. Obviously, $f(x)<a \; \forall x \in A_b$.
Again, by definition of $f$, there are at most countably many $b \in (-\infty,a)$ such that $f(x)=b$ for some $x \in E \;$. In other words, the set $I=\{b \in (-\infty,a): f(x)=b \;$ for some $x \in E \}$ is countable. So, we can write
$f^{-1}(-\infty,a) = f^{-1}\bigcup_{b \in I}\{b\}=\bigcup_{b \in I}f^{-1}\{b\}=\bigcup_{b \in I}A_b \in \mathcal{E}$ where $\bigcup_{b \in I}A_b \in \mathcal{E}$ holds because the countable union of measurable sets is measurable.
In summary, I proved that $f^{-1}(-\infty,a) \in \mathcal{E} \; \forall a \in \mathbb{R} \;$ because it is a countable union of a countable union of measurable sets.
I am not sure my solution is correct or if any of my steps lack some intermediate steps. Any correction would be much appreciated!
There is a more direct route.
For a fixed $a\in\mathbb R$ define index set $I_a\subseteq\mathbb N_+$ by stating that:$$n\in I_a\iff a_n<a$$ Then - because the sets $A_n$ form a partition - we have:$$f^{-1}(-\infty,a)=\bigcup_{n\in I_a}A_n$$ The RHS is a countable union of measurable sets, hence is measurable.