The Unapologetic Mathematician

Mathematics for the interested outsider

The Jordan Decomposition of an Indefinite Integral

So, after all our setup it shouldn’t be surprising that we take an integrable function f and define its indefinite integral:

\displaystyle\nu(E)=\int\limits_E f\,d\mu

Now, as we’ve pointed out, this will be a measure so long as f is a.e. non-negative. But now if f is any integrable function at all, the indefinite integral \nu is a finite signed measure.

Let’s give a Hahn decomposition corresponding to \nu. I say that if we set

\displaystyle\begin{aligned}A&=\{x\in X\vert f(x)\geq0\}\\B&=\{x\in X\vert f(x)<0\}\end{aligned}

then A is positive, B is negative, and X=A\uplus B is a Hahn decomposition. Indeed, we know that A and B are measurable. Thus if E is measurable, then E\cap A is measurable, and we find

\displaystyle\nu(E\cap A)=\int\limits_{E\cap A}f\,d\mu=\int\limits_E f\chi_A\,d\mu\geq0

since f\chi_A=f^+\geq0 everywhere. Similarly, we verify that B is negative.

Now we can use this to find the Jordan decomposition of \nu. We define

\displaystyle\begin{aligned}\nu^+(E)=\nu(E\cap A)&=\int\limits_{E\cap A}f\,d\mu=\int\limits_Ef^+\,d\mu\\\nu^-(E)=\nu(E\cap B)&=\int\limits_{E\cap B}f\,d\mu=-\int\limits_Ef^-\,d\mu\end{aligned}

That is, the upper variation of \nu is the indefinite integral of the positive part of f, while the lower variation of \nu is the indefinite integral of the negative part of f. And then we can calculate

\displaystyle\lvert\nu\rvert(E)=\nu^+(E)+\nu^-(E)=\int\limits_Ef^+\,d\mu+\int\limits_Ef^-\,d\mu=\int\limits_Ef^++f^-\,d\mu=\int\limits_E\lvert f\rvert\,d\mu

The total variation of \nu is the indefinite integral of the absolute value of f.

June 29, 2010 Posted by | Analysis, Measure Theory | 1 Comment



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