The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


  1. […] 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

  2. […] 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

  3. […] 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 […]

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