This is probably simple, but I'm uncertain.
Consider function $f$ to be measurable.
Define $E_n = \{f\geq\frac{1}{n}\}$. Then, for $E = \bigcup_n E_n$. Is $E = \{f > 0\}$ or $E = \{f \geq 0\}$? I think $E = \{f > 0\}$ since $\forall n$, $x\in E_n$ gives $f(x)\geq \frac{1}{n} > 0$.
Similarly, if $E_n = \{f\leq n\}$. Then, $E = \bigcup_n E_n = \{f < \infty\}$ where $f$ not infinite on E.
Are these two conclusions correct?
As mentioned in the comments by @geetha290krm, the conclusions are correct. To elaborate just a bit more, if $f$ is constantly zero, all of the $E_n$ are empty. So then $\bigcup_n E_n$ is empty, but $\{ x : f(x) = 0 \} = \mathbb R$.
For the second bit, if you are allowing for functions to take on value $-\infty$, then you could simply look at $E_n = \{ x : |f(x)| \leq n \}$ or so to obtain the set on which $f$ does not take on any infinite values.