Jordan Decompositions

It’s not too hard to construct examples showing that Hahn decompositions for a signed measure, though they exist, are not unique. But if we have two of them — $X=A_1\uplus B_1=A2\uplus B_2$ — there’s something we can show to be unique. For every measurable set $E$ we have $\mu(E\cap A_1)=\mu(E\cap A_2)$ and $\mu(E\cap B_1)=\mu(E\cap B_2)$ .

Indeed, it’s easy to see that $E\cap(A_1\setminus A_2)\subseteq E\cap A_1$ , so (since $A_1$ is positive) $\mu(E\cap(A_1\setminus A_2))\geq0$ . But we can also see that $E\cap(A_1\setminus A_2)\subseteq E\cap B_2$ , so (since $B_2$ is negative) $\mu(E\cap(A_1\setminus A_2))\leq0$ . And so $\mu(E\cap(A_1\setminus A_2))=0$ , and similarly $\mu(E\cap(A_2\setminus A_1))=0$ . We then see that

$\displaystyle\mu(E\cap A_1)=\mu(E\cap(A_1\cup A_2))=\mu(E\cap A_2)$

The proof that $\mu(E\cap B_1)=\mu(E\cap B_2)$ is similar.

We can thus unambiguously define two set functions on the class of all measurable sets

$\displaystyle\begin{aligned}\mu^+(E)&=\mu(E\cap A)\\\mu^-(E)&=-\mu(E\cap B)\end{aligned}$

for any Hahn decomposition $X=A\uplus B$ . We call these the “upper variation” and “lower variation”, respectively, of $\mu$ . We can also define a set function

$\displaystyle\lvert\mu\rvert(E)=\mu^+(E)+\mu^-(E)$

called the “total variation” of $\mu$ . It should be noted that $\lvert\mu\rvert(E)$ and $\lvert\mu(E)\rvert$ have almost nothing to do with each other.

Now, I say that each of these set functions — $\mu^+$ , $\mu^-$ , and $\lvert\mu\rvert$ — is a measure, and that $\mu(E)=\mu^+(E)-\mu^-(E)$ . If $\mu$ is (totally) finite or $\sigma$ -finite, then so are $\mu^+$ and $\mu^-$ , and at least one of them will always be finite.

Each of these variations is clearly non-negative, and countable additivity is also clear from the definitions. For example, given a pairwise-disjoint sequence $\{E_n\}$ we find

$\displaystyle\begin{aligned}\mu^+\left(\biguplus\limits_{n=1}^\infty E_n\right)&=\mu\left(\left(\biguplus\limits_{n=1}^\infty E_n\right)\cap A\right)\\&=\mu\left(\biguplus\limits_{n=1}^\infty(E_n\cap A)\right)\\&=\sum\limits_{n=1}^\infty\mu(E_n\cap A)\\&=\sum\limits_{n=1}^\infty\mu^+(E_n)\end{aligned}$

and similarly for $\mu^-$ and $\lvert\mu\rvert$ . Thus each one is a measure. The equation $mu=\mu^+-\mu^-$ is clear from the definitions. The fact that $\mu$ takes at most one of $\pm\infty$ implies that one of $\mu^\pm$ is finite. Finally, if every measurable set (say, $E$ ) is a countable union of $\mu$ -finite sets (say, $E=\biguplus E_n$ ), then $E\cap A$ is the countable union of the $E_n\cap A$ , and so $\lvert\mu^+(E_n)\rvert=\lvert\mu(E_n\cap A)\rvert<\infty$ as well, and similarly for $\mu^-(E_n)$ . Thus $\mu^+$ and $\mu^-$ are $\sigma$ -finite.

We see that every signed measure $\mu$ can be written as the difference of two measures, one of which is finite. The representation $\mu=\mu^+-\mu^-$ of a signed measure as the difference between its upper and lower variations is called the “Jordan decomposition” of $\mu$ .

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June 25, 2010 - Posted by John Armstrong | Analysis, Measure Theory

3 Comments »

[...] it so happens that we can define a norm on this space. Indeed, taking the Jordan decomposition, we must have both and , and thus . We define , and use this as our norm. It’s [...]

Pingback by The Banach Space of Totally Finite Signed Measures « The Unapologetic Mathematician | June 28, 2010 | Reply
[...] we can use this to find the Jordan decomposition of . We [...]

Pingback by The Jordan Decomposition of an Indefinite Integral « The Unapologetic Mathematician | June 29, 2010 | Reply
[...] with Respect to a Signed Measure If is a signed measure then we know that the total variation is a measure. It then makes sense to discuss whether or not a measurable function is integrable [...]

Pingback by Integration with Respect to a Signed Measure « The Unapologetic Mathematician | June 30, 2010 | Reply

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