The Unapologetic Mathematician

Mathematics for the interested outsider

Singularity

Another relation between signed measures besides absolute continuity — indeed, in a sense the opposite of absolute continuity — is singularity. We say that two signed measures \mu and \nu are “mutually singular” and write \mu\perp\nu if there exists a partition of X into two sets A\uplus B=X so that for every measurable set E the intersections A\cap E and B\cap E are measurable, and

\displaystyle\lvert\mu\rvert(A\cap E)=0=\lvert\nu\lvert(B\cap E)

We sometimes just say that \mu and \nu are singular, or that (despite the symmetry of the definition) “\nu is singular with respect to \mu“, or vice versa.

In a manner of speaking, if \mu and \nu are mutually singular then all of the sets that give \mu a nonzero value are contained in B, while all of the sets that give \nu a nonzero value are contained in A, and the two never touch. In contradistinction to absolute continuity, not only does the vanishing of \lvert\mu\rvert not imply the vanishing of \lvert\nu\rvert, but if we pare away portions of a set for which \lvert\nu\rvert gives zero measure then what remains — essentially the only sets for which \lvert\nu\rvert doesn’t automatically vanish — is necessarily a set for which \lvert\mu\rvert does vanish. Another way to see this is to notice that if \mu and \nu are signed measures with both \nu\ll\mu and \nu\perp\mu, then we must necessarily have \nu=0; singularity says that \nu must vanish on any set E with \lvert\mu\rvert(E)\neq0, and absolute continuity says \nu must vanish on any set E with \lvert\mu\rvert(E)=0.

As a quick and easy example, let \mu^+ and \mu^- be the Jordan decomposition of a signed measure \mu. Then a Hahn decomposition for \mu gives exactly such a partition X=A\uplus B showing that \mu^+\perp\mu^-.

One interesting thing is that singular measures can be added. That is, if \nu_1 and \nu_2 are both singular with respect to \mu, then (\nu_1+\nu_2)\perp\mu. Indeed, let X=A_1\uplus B_1 and X=A_2\uplus B_2 be decompositions showing that \nu_1\perp\mu and \nu_2\perp\mu, respectively. That is, for any measurable set E we have

\displaystyle\begin{aligned}\nu_1(A_1\cap E)&=0\\\mu(B_1\cap E)&=0\\\nu_2(A_2\cap E)&=0\\\mu(B_2\cap E)&=0\end{aligned}

Then we can write

\displaystyle X=(A_1\cap A_2)\uplus\left((A_1\cap B_2)\uplus(A_2\cap B_1)\uplus(B_1\cap B_2)\right)

It’s easy to check that \nu_1+\nu_2 must vanish on measurable subsets of A_1\cap A_2, and that \mu must vanish on measurable subsets of the remainder of X.

July 5, 2010 Posted by | Analysis, Measure Theory | 1 Comment