Does $f$ measurable $\iff$ $|f|$ measurable?

Question

Does $f$ measurable $\iff$ $|f|$ measurable ?

$\implies $ is clearly true since $$|f|^{-1}((-\infty ,\alpha))=f^{-1}(]-\alpha,\alpha[)$$ and thus $|f|^{-1}$ is measurable.

But for the reciprocal I have doubt. I would says yes if $f$ measurable imply $f^+:= f\vee 0$ and $f^{-}:= f\wedge 0$ measurable, but it sounds wrong to me. Therefore I think it's wrong but I can't find a counter example.

Answer 1

Let $A$ be a non-measurable subset of $\mathbb{R}$, and let $f$ be the function $$f(x) = \begin{cases} 1\text{ if }x\in A\\ -1\text{ if }x\notin A\end{cases}$$

Then $f$ is not measurable ($f^{-1}[\{1\}] = A$) but $|f|$ is measurable (it's the constant function $1$).

Answer 2

A counterexample for $\Leftarrow$:

Take $A$ not measurable and define $f(x) = \begin{cases} 1 & ; x \in A \\ -1 & ; x \notin A \end{cases}$ . $f$ is not measurable but $|f|=1$ is...

Answer 3

I know this question is old, but here's another example. Consider measurable space:

$(X, \mathcal{F}, P)$,

where $X = \{1, 2\}$, $\mathcal{F}=\{\emptyset, X\}$ and $P$ is an arbitrary measure with domain $\mathcal{F}$. Define:

$f(1) = -1$ and $f(2) = 1$.

Now $|f(x)|$ is measurable and $f(x)$ isn't, considering $|f|^{-1}\{(-\infty, c)\} \in \mathcal{F}$, for all real numbers $c$, but $f^{-1}\{(-\infty, 0)\} \not\in \mathcal{F}$.