Polar decomposition of a functional in $C(\Gamma)?$ I'm reading Takesaki's first volume and in the proof of IV.7.3 (chapter 4, proposition 7.3) the following situation happens:
$\Gamma$ is a compact Hausdorff space, and $\mu\in M(\Gamma)$ a complex Radon measure on $\Gamma$. Then Takesaki considers $|\mu|$, which he defines as "the absolute value of the complex measure $\mu$ in the sense of polar decomposition in the conjugate space $C(\Gamma)^*$ of the $C^*$-algebra $C(\Gamma)$."
What does this mean?
Of course, $\mu$ corresponds to a unique functional $\varphi: C(\Gamma)\to \mathbb{C}$ such that
$$\int_\Gamma fd \mu = \varphi(f), \quad f \in C(\Gamma).$$
However, this begs the question: what is the polar decomposition of an element $\varphi$?
I know that it makes sense to consider the polar decomposition of normal functionals in a von Neumann algebra, so perhaps what is going on is that Takesaki identifies $C(\Gamma)^* = (C(\Gamma)^{**})_*$, i.e. $C(\Gamma)^*$ are the normal functionals on the enveloping von Neumann algebra of $C(\Gamma)$ and then he considers the polar decomposition for normal functionals?
If so, does this agree with the usual definition of total variation of a complex measure?
 A: For the Polar Decomposition you need your functional to be normal and you need a von Neumann algebra, so $C(\Gamma)^{**}$ is the natural place to do so.
And yes, the absolute value is the total variation in the sense of a complex measure. Namely, you have $\varphi=u\omega$, which means $\varphi(x)=\omega(xu)$ for all $x$. Given a Borel set $\Delta\subset\Gamma$
\begin{align}
|\mu|(\Delta)
&=\sup\Big\{\sum_{E\in\pi}|\mu(E)|:\ \pi\ \text{ partition of }\Delta\Big\}\\[0.3cm]
&=\sup\Big\{\sum_{E\in\pi}\Big|\int_\Gamma 1_E\,d\mu|:\ \pi\ \text{ partition of }\Delta\Big\}\\[0.3cm]
&=\sup\Big\{\sum_{E\in\pi}|\varphi(1_E)|:\ \pi\ \text{ partition of }\Delta\Big\}\\[0.3cm]
&=\sup\Big\{\sum_{E\in\pi}|\omega(1_E\,u)|:\ \pi\ \text{ partition of }\Delta\Big\}\\[0.3cm]
&\leq\sup\Big\{\sum_{E\in\pi}\omega(1_E):\ \pi\ \text{ partition of }\Delta\Big\}\\[0.3cm]
&=\omega(1_\Delta).
\end{align}
The key inequality above is
$$
|\varphi(1_E)|=|\omega(1_E\,u)|=|\omega(1_E\,1_Eu)|\leq\omega(1_E)^{1/2}\omega(u^*1_Eu)^{1/2}=\omega(1_E)^{1/2}\omega(1_E|u|^21_E)^{1/2}\leq\omega(1_E). 
$$
On the other hand, there exists $h$ measurable, with $|h|=1$, and $d\mu=h\,d|\mu|$ (cfr. Rudin Real and Complex Analysis, Theorem 6.6.12). Then
$$
\omega(1_\Delta)=\varphi(1_\Delta u^*)=\int_\Gamma 1_\Delta\,u^*\,h\,d|\mu|\leq\int_\Gamma 1_\Delta\,d|\mu|=|\mu|(\Delta). 
$$
So
$$
|\mu|(\Delta)=\omega(1_\Delta). 
$$
It then follows from the usual theorems in measure theory that
$$
\omega(f)=\int_\Gamma f\,d|\mu|,\qquad f\in C(\Gamma). 
$$
