The absolute value of a bounded linear functional on a C*-algebra Let us consider the commutative C*-algebra $C_0(\Omega)$. Let $\mu$ be a complex Radon measure on $\Omega$. By the Riesz representation theorem, $\mu$ may be considered as a bounded linear functional on $C_0(\Omega)$. The polar decomposition says that there is a partial isometry $u$ in the W*-algebra $C_0(\Omega)^{**}$ with $\mu=u|\mu|$. 
Question: Is $|\mu|$ just the total variation of $\mu$?
 A: Yes. The measure $\mu$ is absolutely continuous with respect to the total variation $|\mu|$, so by the Radon-Nikodym theorem there exists a function $u \in \mathrm{L}^1(|\mu|)$ such that $\mu = u |\mu|$. The easiest way to identify $\mathrm{L}^1(|\mu|)$ with a subspace of $\mathrm{C}_0(\Omega)^{**}$ is probably by using the Lebesgue decomposition of $\mathrm{C}_0(\Omega)$ into absolutely continuous and singular parts with respect to $|\mu|$. You can also verify that multiplication of a function by a measure agrees with the natural dual module action of $\mathrm{C}_0(\Omega)^{**}$ on $\mathrm{C}_0(\Omega)^*$.
The polar decomposition applied to $\mu$ as a normal linear functional on $\mathrm{C}_0(\Omega)^{**}$ gives a positive $|\mu| \in \mathrm{C}_0(\Omega)^*$ and $v \in \mathrm{C}_0(\Omega)^{**}$ such that $\mu = v |\mu|$. The uniqueness of $|\mu|$ holds unconditionally, but $v$ is only unique up to the condition that $v^*v$ is the support projection of $|\mu|$. You need to show that you can identify the support projections of elements in $\mathrm{C}_0(\Omega)^*$ and $\mathrm{C}_0(\Omega)^{**}$ with subsets of $\Omega$ and that this agrees with the usual notion of support for measures on locally compact Hausdorff spaces. Since the measure-theoretic Radon-Nikodym derivative of $\mu$ with respect to $|\mu|$ obviously has the same support as $\mu$, you can apply the uniqueness result.
