# The Unapologetic Mathematician

## 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 »

1. […] define to be the indefinite integral of . We can tell that the total variation is the Lebegue measure itself, since . Thus if then we can easily […]

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