# The Unapologetic Mathematician

## 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 $X$.