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Let $(X, \mathcal X)$ be a measurable space. Let $\mu$ be a complex measure on $X$ and $|\mu|$ its variation. Then $|\mu|$ is a non-negative finite measure. By definition, $|\mu(B)| \le |\mu| (B)$ for all $B \in \mathcal X$.

Let $\mu_1, \mu_2$ be the real and imaginary parts of $\mu$. Then $\mu_1, \mu_2$ are finite signed measures such that $\mu = \mu_1 + i \mu_2$. Let $(\mu_1^+, \mu_1^-)$ and $(\mu_2^+, \mu_2^-)$ be the Jordan decompositions of $\mu_1, \mu_2$ respectively. Then $\mu_1^+, \mu_1^-, \mu_2^+, \mu_2^-$ are non-negative finite measures such that $\mu_1 = \mu_1^+ - \mu_1^-$ and $\mu_2 = \mu_2^+ - \mu_2^-$.

Integration w.r.t. $\mu$ is defined as follows.

  • If $f:X \to \mathbb R$ is measurable then $$ \begin{align} \int_X f \mathrm d \mu &:= \int_X f \mathrm d \mu_1 + i \int_X f \mathrm d \mu_2 \\ &:= \left [ \int_X f \mathrm d \mu_1^+ - \int_X f \mathrm d \mu_1^- \right ] + \left [ \int_X f \mathrm d \mu_2^+ - \int_X f \mathrm d \mu_2^- \right ], \quad (\star) \end{align} $$ provided that each integral in $(\star)$ is well-defined.
  • If $f:X \to \mathbb C$ is measurable then $$ \int_X f \mathrm d \mu := \int_X (\operatorname{Re} f) \mathrm d \mu + i \int_X (\operatorname{Im} f) \mathrm d \mu. $$

In a proof of this result, I appealed to below inequality many times.

Theorem: If $f:X \to \mathbb R$ measurable bounded, then $$ \left | \int_X f \mathrm d \mu \right | \le \int_X |f| \mathrm d |\mu| $$

As such, I would like to prove it. Could you have a check on my attempt?

Proof: For convenience, let $\alpha$ be the value of the LHS and $\beta$ that of the RHS.

  1. Let $f = 1_B$ with $B \in \mathcal X$. So $f$ is a characteristic function.

We have $\alpha = |\mu(B)|$ and $\beta = |\mu| (B)$. The claim then holds.

  1. Let $f = \sum_{i=1}^m b_i 1_{B_1}$ with $b_i \in \mathbb R$ and $B_i \in \mathcal X$ such that $B_i \cap B_j \neq \emptyset \iff i=j$. So $f$ is a simple function.

We have $\alpha = |\sum_{i=1}^m b_i \mu(B_i)|$ and $\beta = \sum_{i=1}^m |b_i| \cdot |\mu| (B_i)$. The claim then holds thanks to triangle inequality and (1.)

  1. Let $f:X \to \mathbb R$ be measurable bounded.

There is a sequence $(f_n)$ of simple functions such that $(f_n)$ is uniformly bounded and that $f_n \to f$ pointwise everywhere. By applying DCT for each term in $(\star)$, we have $$ \alpha = \left | \lim_n \int_X f_n \mathrm d \mu \right | = \lim_n \left | \int_X f_n \mathrm d \mu \right | . $$

By (2.), we get $$ \alpha \le \lim_n \int_X |f_n| \mathrm d |\mu|. $$

By DCT again, we have $$ \lim_n \int_X |f_n| \mathrm d |\mu| = \int_X |f| \mathrm d |\mu| = \beta. $$

This completes the proof.

Analyst
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    For a reference, @DavidC.Ullrich gave an elegant proof based on the polar decomposition of a complex measure here. – Analyst Nov 09 '22 at 09:04
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    (+1) Once you have the (real version) Radon-Nikodym theorem at your disposal, it is easy to get a complexificaton of it and see that $d\mu=h,d|\mu|$ with $|h|=1$ using the definition of variation measure. This will simplify your proof considerably. Since you are interested in complexification of operators, I think this may be interesting to you – Mittens Nov 09 '22 at 14:11
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    Also, this Bachellor's thesis has some interesing nuggets – Mittens Nov 09 '22 at 14:15

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