# The Unapologetic Mathematician

## 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 $\mu$.

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

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