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I want to prove that if $\mu$ is a complex measure, then:

$$\left|\int f\,d\mu\right|\le \int |f| \, d|\mu|$$

Where $|\mu|(E)=\sup\{\sum_{n=1}^N |\mu(E_n)|\}$, where $\{E_n\}$ is a measurable partition of E(the total variation). By splitting $\mu$ in a real and imaginary part, and then using the Jordan-decomposition for signed measures. We can write $\mu=\mu_1^+-\mu_1^-+i\mu_2^+-i\mu_2^-$, where all four parts on the left side are finite meassures, and the two first are mutually singular, and the two last are mutually singular.

I tried using the bootstrap argument. And I think I got it to work for characteristic functions, then simple functions, then positive funtions using the MCT, then real functions by splitting up in a positive or negative part.

However the last step is where I am stuck, going from real to complex functions. What worked for real functions by splitting up in a positive or negative part does not seem to work for complex funtions. Because this was the step going from positive to real:

$|\int f^+-f^- \, d\mu|\le|\int f^+\,d\mu|+|\int f^-\,d\nu|$, then using that it holds for positive functions I get $\le \int |f^+\,|d|\mu|+\int|f^-| \, d|\mu| = \int|f| \, d|\mu|$. But what made this work is the fact that:

$|f^-|+|f^+|=|f|$, since if one of them is always zero for a given x. But for complex functions we don't have:

$|\Re f|+|\Im f|=|f|$ So do you have any tips on how to finish the bootstrap-argument?

PS: I do not know if there are many different ways to describe the integral of a function with respect to a complex measure, but my book defines it:

$\int f \, d\mu=\int f \, d\mu_1^+-\int f \, d\mu_1^-+i\int f\,d\mu_2^+-i\int f\,d\mu_2^-$, and then if the function is complex valued you split it up in a complex and negative part etc., but this is no problem since the 4 measures on the RHS are finite positive measures, so it is treated in the ordinary measure theory.

Michael Hardy
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user119615
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2 Answers2

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As it happens, a few months ago I was stuck on exactly this point; I was surprised that what you refer to as "bootstrapping" didn't seem to work. What I finally came up with was something that I suspect is analogous to what Patrick Da Silva said; haven't looked at his answer in detail but it seems similar.

This was because I wanted to do it without the Radon-Nikodym theorem. If you use RN it's very simple:

It's clear from the definitions that $\mu<<|\mu|$. So RN shows that there exists a function $\alpha$ with $d\mu=\alpha\, d|\mu|$, or in other words $$\int f\,d\mu=\int f\alpha\,d|\mu|.$$ It's not hard to show that $|\alpha|=1$ almost everywhere wrt $|\mu|$ (see Details below) and now you're done, using the inequality for positive measures: $$\left|\int f\,d\mu\right|=\left|\int f\alpha\,d|\mu|\right|\le\int|f\alpha|\,d|\mu|=\int|f|\,d|\mu|.$$

Details: First note that there's no need to worry that we might need to prove this inequality first before proving RN. The RN we need is immediate from RN for positive measures. Say $\mu=(\mu_1-\mu_2)+i(\mu_3-\mu_4)$ where $\mu_j\ge0$, $\mu_1$ and $\mu_2$ have disjoint support and $\mu_3$ and $\mu_4$ have disjoint support. Then it's easy to see that $\mu_j<<|\mu|$, so RN for positive measures shows that there exists $\alpha_j$ with $d\mu_j=\alpha_j\,d|\mu|$, and hence there exists $\alpha$ with $d\mu=\alpha\,d|\mu|$.

And it's easy to see that $|\alpha|\le 1$ ae [$|\mu|$]: For any $E$ we have $$\left|\int_E\alpha\,d|\mu|\right|=|\mu(E)|\le|\mu|(E).$$

We don't actually need $|\alpha|\ge1$, but that's easy as well. Say $\lambda<1$ and $|\alpha|\le\lambda$ on $E$. Then the definition of $|\mu|(E)$ in terms of partitions of $E$ shows that $$|\mu|(E)\le\lambda|\mu|(E),$$hence $|\mu|(E)=0$.

David C. Ullrich
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Write $f = Rf + i If \in \mathcal L^1(X,\Sigma, \mu)$, its real and imaginary parts. It is already clear that $$ \left| \int_X Rf \, d\mu \right| \le \int_X |Rf| \, d|\mu|, \quad \left| \int_X If \, d\mu \right| \le \int_X |If| \, d|\mu|. $$ since this is the case where the function integrated is real. Therefore, $$ \left| \int_X Rf \, d\mu \right|^2 + \left| \int_X If \, d\mu \right|^2 \le \left| \int_X |Rf| \, d|\mu| \right|^2 + \left| \int_X |If| \, d|\mu| \right|^2 . $$ To finish the proof, it suffices to show that if $\mu$ is a finite positive measure and $f,g \in \mathcal L^1(X,\Sigma,\mu)$ are positive real functions, then $\sqrt{f^2+g^2} \in \mathcal L^1(X,\Sigma,\mu)$ and $$ \sqrt{\left| \int_X f \, d\mu \right|^2 + \left| \int_X g \, d\mu \right|^2} \le \int_X \sqrt{f^2+g^2} \, d\mu. $$ The first part is easy since $\sqrt{f^2+g^2} \le |f|+|g| = f+g \in \mathcal L^1(X,\Sigma,\mu)$. Since we assume $f,g$ positive, we can prove this for simple functions (where this is essentially Cauchy-Schwarz) and then use MCT.

When you've proved this, $$ \left| \int_X f \, d\mu \right| = \sqrt{\left| \int_X Rf \, d\mu \right|^2 + \left| \int_X If \, d\mu \right|^2} \le \sqrt{\left| \int_X Rf \, d|\mu| \right|^2 + \left| \int_X If \, d|\mu| \right|^2} \le \int_X \sqrt{Rf^2 + If^2} d|\mu| = \int_X |f| d|\mu|. $$

Hope that helps,

Patrick Da Silva
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