Problem: Give example where $\left|f\right|$ is $\mathcal{A}$-measurable but $f$ is not $\mathcal{A}$-measurable. This problem is related to the following two:
If |f| is (lebesgue) measurable, is f measurable?
But I think my stated problem is more simple. Is my reasoning (below) correct (even though it is sloppy)? Is it possible to make my proof more "rigorous"?
My attempt: define $E_{a}=\left\{ x:-a\le x\le a\right\} $. Let $\mathcal{A}$ be the smallest $\sigma$-algebra generated from the collection $\left\{ E_{a}:a\in\mathbb{R}\right\} $. Then $\left(\mathbb{R},\mathcal{A}\right)$ is a measurable space. Define $f:\mathbb{R}\rightarrow\mathbb{R}$ to be $f(x)=x$. Consider the set $G=(-\infty,b]$ for a specific $b<0$ (say). I now try to prove that $G\not\in\mathcal{A}$. Let's see what the intersection-criteria of a $\sigma$-algebra can do.
For any collection $E_{a_{1}},E_{a_{2}},\ldots,E_{a_{\infty}}$, where $a_{i}\in\mathbb{R}$, consider that
$\bigcup_{i=1}^{\infty}E_{a_{i}}=E_{\sup a_{i}}$
Clearly, $G\ne E_{\sup a_{i}}$ for any sequence $\left\{ a_{i}\right\} $.
Similarly, the $E_{a}^{c}$ won't help, as any countable union or intersection of $E_{a_{i}}$ and $E_{b_{i}}^{C}$ will either be the empty set or will have a real number on both sides of 0, so it is impossible that it could equal $G$. Thus $G\not\in\mathcal{A}$ and so $f$ is not $\mathcal{A}$-measurable.
There is a shortcut for proving that sets like $(-\infty,b]$ will not belong to the $\sigma$-algebra generated by the sets of the form $[-a,a]$.
Let $\mathcal V\subset\wp(\mathbb R)$ denote the collection of sets $S$ that satisfy $x\in S\implies-x\in S$ for every $x\in\mathbb R$.
Then $\mathcal V$ is evidently a $\sigma$-algebra.
Also evident is that $\mathcal V$ contains all sets of the form $[-a,a]$.
This together allows the conclusion that $\mathcal A\subseteq\mathcal V$ where $\mathcal A$ denotes the smallest $\sigma$-algebra containing these sets.
Finally it is evident that $(-\infty,b]\notin\mathcal V$ hence $(-\infty,b]\notin\mathcal A$.