# Square root of a complex measurable function

Let $f:\mathbb{R}\to \mathbb{C}$ be a Lebesgue measurable function. For each $z\in \mathbb{C}$ let $Arg (z)$ be the principal argument of $z$ (define it to be $0$ if $z=0$). Define $g(x)= \sqrt{|f(x)|}exp(\frac{1}{2} i Arg (f(x))$. Then $g(x)$ is measurable and $(g(x))^2=f(x)$.

I am sure the above statement is correct, but just wanted to confirm with you all. The only part of concern is, of course, whether the function $exp(\frac{1}{2} i Arg (f(x))$ is measurable.

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$\exp({1 \over 2}i(Arg(f(x)))$ can be written as $h(g(f(x)))$, where $h(z) = e^{{1 \over 2} iz}$ and $g(z) = Arg(z)$. By Rudin thm 1.12d) for example, a measurable function followed by a Borel measurable function is measurable. So it suffices to show $g$ and $h$ are Borel measurable. For $h(z)$ this follows from the fact that $h$ is continuous. For $g(z)$, regardless of how you define $Arg(z)$, the set $\{z: a < Arg(z) < b\}$ is the union of at most two wedges which are Borel sets. So $\{z: g(z) \in (a,b) \}$ is a Borel set. This means $g(z)$ is Borel measurable as well.
@Zaricause. A Lebesgue measurable function $f$ followed by a Borel measurable function $Arg$ need not measurable, I think. So $Arg(F(x)$ being measurable is in question. – TCL Nov 20 '10 at 16:29
I showed $Arg f(x)$ is measurable by showing the inverse image of an open interval is measurable... this is the definition of measurable. To see that Lebesgue measurable followed by Borel measurable actually is measurable, I refer you to Rudin Thm 1.12d). – Zarrax Nov 20 '10 at 17:14