# If $f$ is real-valued bounded measurable and $\mu$ a complex measure, then $\left | \int_X f \mathrm d \mu \right | \le \int_X |f| \mathrm d |\mu|$

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.

• For a reference, @DavidC.Ullrich gave an elegant proof based on the polar decomposition of a complex measure here. Nov 9, 2022 at 9:04
• (+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 Nov 9, 2022 at 14:11
• Also, this Bachellor's thesis has some interesing nuggets Nov 9, 2022 at 14:15