In Folland's "Real Analysis", two signed measures $\mu,\nu$ on a measurable space $(X,\mathcal{M})$ are said to be mutually singular if there is a decomposition of $X=E\cup F$ so that $E$ is null for $\mu$ and $F$ is null for $\nu,$ and he denotes it by $\mu\bot\nu.$
The notation "$\bot$" caught my interest, since we often use it to represent perpendicularity in inner product space. So if we set
$$V=\left\{\text{all finite signed measures on a measurable space}~(X,\mathcal{M})\right\}$$
it's clear that $V$ is a real vector space. So my question is: can we put an inner product $\langle\cdot,\cdot\rangle$ on $V$ such that $\langle\mu,\nu \rangle=0$ exactly when $\mu,\nu$ are mutually singular?
(If this is true, then the Radon-Nikodym theorem just states that for any fixed $\mu\in V,$ $V$ has an orthogonal decomposition of $MS_{\mu}$ and $AC_{\mu},$ where $MS_{\mu}$ consists of all $\nu\in V$ that is mutually singular to $\mu$ and $AC_{\mu}$ consists of all $\nu\in V$ that's absolutely continuous to $\mu,$ which is really interesting.)
Best Answer
No, in general, the relation "mutual singularity" does not coincide with the orthogonality relation stemming from any inner product.
For instance, endow $\{0,1\}$ with its power set as $\sigma$-algebra. Let $\mu$ be the measure that assigns $1/2$ to each of the singletons $0,1$. Then no non-zero measure is singular to $\mu$. But in a two-dimensional inner product space every vector has non-zero orthogonal complement.
If you're looking for an abstract generalization of mutual singularity of measures, the right notion is "disjointness" in vector lattices.