I hope you can help me with this question:
Let ($\Omega,\Sigma,\mu$) be a measurable space. Let $f:\Omega\rightarrow \mathbb{R}$ be a real-function. I would like to know: If $\vert f \vert$ is $\mu$-measurable, is $f$ $\mu$-measurable too?
Thanks in advance!
In general, an easy counterexample can be constructed whenever there's a non-measurable set, i.e. if there exists $F \in \mathcal{P}( \Omega ) \setminus \Sigma$. Then setting $$f(x) = \begin{cases} 1 & x \in F \\ - 1 & x \not \in F \end{cases} ,$$ we have a function whose absolute value is constant (thus measurable), but is itself not measurable. So in fact the measurability of $|f|$ implies the measurability of $f$ iff every function is measurable anyways (i.e. $\Sigma = \mathcal{P}(\Omega)$).
No. Let $\Omega = \{\omega_1, \omega_2 \}$, and let $\Sigma = \{\emptyset, \Omega \}$. Let $f(\omega_1) = 1$ and $f(\omega_2) = -1$. Then $|f| = 1$ is constant and hence $\Sigma$-measurable. But $\{f \leq -1 \} = \{\omega_2 \} \notin \Sigma$, so $f$ is not $\Sigma$-measurable.
(One normally writes "$\Sigma$-measurable" because the measurability of $f$ depends on the $\sigma$-algebra associated with $\Omega$, not the measure $\mu$.)