I am trying to prove the following:

(I) Let $f:X\rightarrow \mathbb{C}$ (i.e. complex space) and $f=u+iv$ where $u$ is the real part and $v$ is the imaginary part. If $f$ is measurable, then the magnitude of $f$ given as $|f|$ is measurable.

(II) If $f$ is measurable then the sign function of $f$ given as

$$sign(f)=\begin{cases}\frac{f}{|f|},&f\neq 0,\\0,&else,\end{cases}$$

is also measurable.

Thanks in advance

  • $\begingroup$ Just prove that if $f:X\to \mathbb{C}$ is measurable and $\varphi : \mathbb{C}\to \mathbb{C}$ continuous, then $\varphi\circ f$ is measurable. $\endgroup$ – Prahlad Vaidyanathan Sep 23 '15 at 1:31
  • $\begingroup$ I am not sure how this would prove this. I was thinking of using the definition of the magnituge |f|=(f^+) + (f^-) and take it from there but not sure how? $\endgroup$ – Ahmad Abdulrahman Sep 23 '15 at 1:36
  • $\begingroup$ What is your definition of measurability of a function? $\endgroup$ – Prahlad Vaidyanathan Sep 23 '15 at 1:37
  • $\begingroup$ f is measurable if {x:f(x)>a} is in the sigma algebra of the measure space for all real numbers a. $\endgroup$ – Ahmad Abdulrahman Sep 23 '15 at 1:39
  • $\begingroup$ Well, this is the set $f^{-1}((a,\infty))$. Can you prove that this is equivalent to saying that $f^{-1}(U)$ is measurable for any open set $U \subset \mathbb{C}$? From that, the statement I made above follows quite easily. $\endgroup$ – Prahlad Vaidyanathan Sep 23 '15 at 1:42

For both parts, you can use the fact that composition of measurable functions gives a measurable function*. With that in mind, for part 1, you only have to show that the magnitude function $z \mapsto |z|$ from $\mathbb{C} \to \mathbb{R}$ is Borel measurable. This function is continuous, and is therefore Borel measurable. For part 2, we need to show that $sign$ is Borel measurable. Use the fact that we can write $sign$ in countably many (even finitely many) continuous pieces (with Borel domains).

*As per a comment, we need to be careful about $\sigma$-algebras a little bit. If we are considering $f \circ g$ with, e.g., $g:X \to \mathbb{R}$ and $f:\mathbb{R} \to \mathbb{R}$, we need to be talking about the same $\sigma$-algebra for the domain of $f$ and codomain of $g$. To say that $g$ is measurable usually means (as it does in your definition) that we are talking about the Borel $\sigma$-algebra on $\mathbb{R}$, so we should be making sure that $f$ is a Borel-measurable function, also, in that the inverse image of each $\{x : x > a\}$ is a Borel set. The approaches I outline do show measurability in this stronger sense.

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  • $\begingroup$ What you say is not true: math.stackexchange.com/questions/283443/… $\endgroup$ – Prahlad Vaidyanathan Sep 23 '15 at 5:34
  • $\begingroup$ @PrahladVaidyanathan Added caveat to clarify we need to show that the outer function in the composition is Borel-measurable. $\endgroup$ – user231101 Sep 23 '15 at 10:30

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