Product and signed measures · Definition

Hahn–Jordan decomposition

measure theory and integrationproduct and signed measures

Every signed measure can be split into positive and negative parts using a Hahn set. This idea links signed measures with ordinary positive measures.

Definition

The Hahn–Jordan decomposition splits a signed measure \(\nu\) into the difference of two mutually singular positive measures, \(\nu = \nu^+ - \nu^-\).

\[\nu = \nu^+ - \nu^-,\quad \nu^+ \perp \nu^-\]

If \(\nu\) is a signed measure on a \(\sigma\)-algebra \(\mathcal{M}\), there exists a Hahn set \(P\) such that \(\nu(E)\ge 0\) for every measurable \(E\subseteq P\) and \(\nu(E)\le 0\) for every measurable \(E\subseteq P^c\). Using it, define \(\nu^+(E)=\nu(E\cap P)\) and \(\nu^-(E)=-\nu(E\cap P^c)\), which are mutually singular positive measures.

Let \(X=\mathbb{R}\) with \(\nu(E)=\int_E f\,dm\), where \(f(x)=\sin x\). The set \(P=\{x: \sin x\ge 0\}\) is a Hahn set, and the decomposition gives \(\nu^+(E)=\int_{E\cap P}\sin x\,dm\) and \(\nu^-(E)=-\int_{E\cap P^c}\sin x\,dm\).

A typical error is thinking that \(\nu^+\) and \(\nu^-\) are simply \(\nu\) restricted to two arbitrary halves of the space. The Hahn set must be chosen so that \(\nu\) is nonnegative on \(P\) and nonpositive on \(P^c\); any other partition does not give the correct decomposition.

Think about it

Why must \(\nu^+\) and \(\nu^-\) be mutually singular and not merely positive?

Without singularity, \(\nu\) would admit many decompositions \(\mu_1-\mu_2\), such as \(\nu^+ + \lambda\) and \(\nu^- + \lambda\) for any positive measure \(\lambda\); singularity fixes a unique split.

Exercises 10

References

  • Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd edition, Wiley, 1999.
  • H. L. Royden, P. M. Fitzpatrick, Real Analysis, 4th edition, Pearson, 2010.