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Here (Section "Integration in respect to a complex measure"), they say that:

Another approach is to not develop a theory of integration from scratch, but rather use the already available concept of integral of a real-valued function with respect to a non-negative measure. To that end, it is a quick check that the real and imaginary parts μ1 and μ2 of a complex measure μ are finite-valued signed measures. One can apply the Hahn-Jordan decomposition to these measures to split them as $$ \mu_1=\mu_1^+-\mu_1^-$$ and $$ \mu_2=\mu_2^+-\mu_2^- $$ where $μ_1^+, μ_1^-, μ_2^+, μ_2^-$ are finite-valued non-negative measures (unique in some sense).

Where does the minus sign come from? What about $\mu_1^+ +\mu_1^-$?

EDIT Further they write

Then, for a measurable function f which is real-valued for the moment, one can define $$ \int_X \! f \, d\mu = \left(\int_X \! f \, d\mu_1^+ - \int_X \! f \, d\mu_1^-\right) + i \left(\int_X \! f \, d\mu_2^+ - \int_X \! f \, d\mu_2^-\right) \tag{1} $$ as long as the expression on the right-hand side is defined, that is, all four integrals exist and when adding them up one does not encounter the indeterminate ∞−∞. Given now a complex-valued measurable function, one can integrate its real and imaginary components separately as illustrated above and define, as expected, $$ \int_X \! f \, d\mu = \int_X \! \Re(f) \, d\mu + i \int_X \! \Im(f) \, d\mu. \tag{2} $$

How does (1) related to (2)? Why doesn't something like $[d]\Re(\mu)$ or show up?

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(1) is the definition of integration of a real valued function with respect to the complex measure $\mu$ -- in terms of four integrals with respect to positive measures (which I assume is familiar to you) (2) goes on to define integrals of complex valued functions with respect to the complex measured by splitting the function $f$ in its real and imaginary parts –  mrf Mar 12 '12 at 22:31
    
@mf Thanks so far, but I'm not familar with positive measure. Do you have a good reference? –  draks ... Mar 12 '12 at 22:36
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Oh, sorry, it seemed like a natural assumption. I wouldn't recommend trying to understand complex measures without doing positive measures first. There are several dozen if not hundreds of decent textbooks in measure theory and integration theory. Not sure I can single out one in particular. You might look at math.stackexchange.com/questions/71418/… –  mrf Mar 12 '12 at 22:42
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2 Answers

up vote 1 down vote accepted

Perhaps an example may help. Consider the complex measure $d\mu = (\cos(x)+ i \sin(x))\ dx$ on $[0,2\pi]$. The real part is $d\mu_1 = \cos(x)\ dx$ and the imaginary part is $d\mu_2 = \sin(x)\ dx$. We then split up the real part as $d\mu_1 = d\mu_1^+ - d\mu_1^-$ where $d\mu_1^+ = \rho_1^+(x)\ dx$, $d\mu_1^- = \rho_1^-(x)\ dx$, $\rho_1^+(x) = \max(\cos(x),0)$ and $\rho_1^-(x) = -\min(\cos(x),0)$. Similarly $d\mu_2 = d\mu_2^+ - d\mu_2^-$. Note that $d\mu_1^+$, $d\mu_1^-$, $d\mu_2^+$ and $d\mu_2^-$ are all positive measures (the whole point of this exercise being that we already know how to integrate with a positive measure, so this construction lets us define integration with a complex measure in terms of integrations with positive measures). That is, $\int_X f\ d\mu$ is defined to be $\int_X f\ d\mu_1^+ - \int_X f\ d\mu_1^- + i \int_X f \ d\mu_2^+ - i \int_X f \ d\mu_2^-$.

Actually we know how to integrate real-valued functions with a positive measure, but what about complex-values functions? That's what (2) is about. If $F$ is a complex-valued function, you split it up into its real and imaginary parts, say $F = g + i h$ where $g$ and $h$ are real, and then $\int_X F \ d\mu = \int_X g \ d\mu + i \int_X h \ d\mu$. And we just saw in the last paragraph how to integrate the real-valued functions $g$ and $h$ with the complex measure $\mu$, so we're done.

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+1 for very helpful example. You should put it there or link your answer. –  draks ... Mar 12 '12 at 22:54
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In the Hahn decomposition, the usual way is to split the measure into two positive measures, hence the minus sign. In other words, $\mu^-$ is not the negative part of $\mu$, but "minus the negative part".

See Wikipedia: Hahn decomposition theorem

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Thanks so far, but I still don't understand that Wiki page. Is this a typo in there? –  draks ... Mar 12 '12 at 22:26
    
Where is the typo? –  mrf Mar 12 '12 at 22:33
    
"Nowhere, everythings fine" is also an accepted answer (to my comment). –  draks ... Mar 12 '12 at 22:39
    
Ok, it sounded like you had something particular in mind. I couldn't find any glaringly obvious typos. –  mrf Mar 12 '12 at 22:44
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