3

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.

Rick
  • 1,707
  • You need to conclude $f$ is measurable from $|f|$ measurable, sin it's not clear how knowing $f$ measurable implies anything would help you prove that $f$ is measurable. – Thomas Andrews Nov 09 '15 at 17:43
  • @ThomasAndrews: I don't understand what you want to say. What is not clear ? The fact that $|f|$ measurable knowing $f$ is measurable ? It comes from the fact that for all $\alpha$ $$|f|^{-1}((-\infty ,\alpha))=f((-\alpha,\alpha))$$ and since $f$ is measurable and that $(-\alpha,\alpha)$ is borel, then $f^{-1}((-\alpha,\alpha))$ is measurable. Therefore, for all $\alpha$, $|f|^{-1}((-\infty ,\alpha)$ is measurable. – Rick Nov 09 '15 at 17:52
  • You wrote "But for the reciprocal I have doubt. I would says yes if ..." The reciprocal, I took to mean the other direction, but maybe you mean the inverse function? The inverse function $f^{-1}$ is never, or rarely, called a reciprocal. – Thomas Andrews Nov 09 '15 at 17:57
  • Ohh, I see. I mean the reciprocal of the proposition (i.e. $|f|$ measurable $\implies f$ measurable). But in fact, thank you for the information about the inverse of the function, I'm sure I would have ask the question in the futur, so it's a good thing that you anticipate this point. Thank you very much :-) – Rick Nov 09 '15 at 18:03
  • That's usually called "the converse," not reciprocal. @Rick – Thomas Andrews Nov 09 '15 at 18:04
  • In fact, does the inverse (if it exist) of a measurable function if measurable ? (thank you for correcting my english, I will know now :-) ) @ThomasAndrews – Rick Nov 09 '15 at 18:05

3 Answers3

9

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 $1$).

Alex Kruckman
  • 76,357
8

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...

Stephan Kulla
  • 7,091
0

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 $f^{-1}\{(-\infty, 0)\} \not\in \mathcal{F}$.

TOMILO87
  • 510