I recently posted this question Trouble proving outer regularity of measure in proof of the multiplication version of the spectral theorem in Dunford & Schwartz where I was not able to confirm the outer regularity of the measure, $\mu$, constructed in the proof of the theorem. I am now beginning to wonder if, in fact, it is not necessarily true that $\mu$ is regular. So I'd like to explore creating a counterexample. Obviously $\mathfrak{H}$ would have to be non-separable. But my repertoire of bounded normal operators on non-separable Hilbert spaces along with knowledge of their spectra and spectral decompositions is extremely limited. What I'm hoping to get to is a maximal collection of mutually orthogonal closed reducing subspaces, $\{\mathfrak{H}_\alpha:\alpha\in M\}$ of $\mathfrak{H}$, for a bounded normal operator $T$, where $M$ is uncountable, and such that $\sigma(T_\alpha)$ ($T_\alpha$ is the restriction of $T$ to $\mathfrak{H}_\alpha$) and the resolution of the identity for $T$ are easy to calculate/describe. My hope is then for an uncountable number of $\alpha\in M$, that I can find a Borel subset, $A$, of $\sigma(T_\alpha)$, which is not relatively open in $\sigma(T_\alpha)$, so that any containing relatively open subset of $\sigma(T_\alpha)$ has nonzero measure with respect to the finite regular measure $\mu_\alpha$ derived from the the resolution of the identity for $T$, as in the theorem.
So, what are some good candidates for
- the non-separable Hilbert space $\mathfrak{H}$
- the bounded normal operator $T$ on $\mathfrak{H}$?
An additional question would be, even if there is a counterexample, is the theorem in Dunford & Schwartz still true? That is does there exist a regular $\mu$, but the measure created in the proof isn't necessarily it? And the final question would be, should I care if $\mu$ is not regular? Is it important for applications?