In Dunford & Schwartz Linear Operators Part II X.5.4, they start with a bounded normal operator $T$ on a (not necessarily separable) Hilbert space $\mathfrak{H}$. In the previous theorem, they proved that there is a maximal collection of mutually orthogonal closed subspaces $\{\mathfrak{H}_\alpha:\alpha\in M\}$ of $\mathfrak{H}$, each of which is reducing for $T$, with the additional property that each $\mathfrak{H}_\alpha$ has an element $\bar{x}_\alpha$ such that the set \begin{equation*} \{p(T_\alpha,T_\alpha^*)\bar{x}_\alpha: p(T_\alpha,T_\alpha^*) \text{ is a polynomial in $T_\alpha$ and $T_\alpha^*$}\} \end{equation*} is dense in $\mathfrak{H}_\alpha$, where $T_\alpha$ is the restriction of $T$ to $\mathfrak{H}_\alpha$. They also show that for each $\alpha\in M$, there is a finite positive regular Borel measure $\mu_\alpha$ on the Borel subsets of $\mathbb{C}$, which vanishes on $\sigma(T_\alpha)^c\supseteq\sigma(T)^c$, and a Hilbert space isomorphism $U$ of $\mathfrak{H}$ onto $\sum_{\alpha\in M} L^2(\mu_\alpha)$ such that for every Borel measurable function, $f$, defined and bounded on $\sigma(T)$, and every $x\in\mathfrak{H}$, and every $\alpha\in M$, $$(U(f(T)x))_\alpha(\lambda)=f(\lambda)(Ux)_\alpha(\lambda),$$ for $\mu_\alpha$-almost all $\lambda$. ($\sum_{\alpha\in M} L^2(\mu_\alpha)$ is the Hilbert space sum of the $L^2(\mu_a)$ Hilbert spaces.)
In the present theorem, they will define a set $S$, a $\sigma$-algebra, $\Sigma$, on $S$ and a regular positive measure $\mu$ on $\Sigma$. There are other parts to the theorem, but where I am having trouble is in the proof of the outer regularity of $\mu$ (inner regularity does not seem to pose any problems). So, what follows are the definitions of $S$, $\Sigma$, and $\mu$.
For each $\alpha\in M$, let $S_\alpha=\sigma(T_\alpha)$. Let $$S=\{(s_\alpha,\alpha):s_\alpha\in S_\alpha,\,\alpha\in M\}.$$ So $S$ is the disjoint union of all of the $S_\alpha$, where they've been "tagged" by $\alpha$ to keep them separate. If $e\subseteq S$, let $$e_\alpha=\{s_\alpha: (s_\alpha,\alpha)\in e\}.$$ I call this a "slice" of $e$. Note that the above notation makes sense, even if $S$ is substituted for $e$. Let $$\Sigma=\{e\subseteq S: e_\alpha\in\mathscr{B}_\alpha\text{ for each $\alpha\in M$}\},$$ where $\mathscr{B}_\alpha$ are the Borel subsets of $S_\alpha$ (which is the same as the Borel subsets of $\mathbb{C}$ which are contained in $S_\alpha$). Let $$\mu(e)=\sum_{\alpha\in M}\mu_\alpha(e_\alpha)\qquad(e\in\Sigma),$$ if the sum converges, and $\infty$ otherwise. To talk about regularity, we need a topology for $S$. We will define a metric. Let \begin{equation*} d((s_\alpha,\alpha),(s_\beta,\beta))= \begin{cases} \lvert s_\alpha-s_\beta\rvert&\text{if $\alpha=\beta$},\\ \lvert\lvert T\rvert\rvert&\text{otherwise}. \end{cases} \end{equation*}
I have no trouble showing that $d$ is a metric on $S$, that for $r\leq\lvert\lvert T\rvert\rvert$, \begin{equation*} B_S((s_\alpha,\alpha),r)=\{(s_\beta,\beta)\in S: d((s_\alpha,\alpha),(s_\beta,\beta))<r\} =(B(s,r) \cap S_\alpha)\times\{\alpha\}, \end{equation*} that $e\subseteq S$ is open if and only if $e_\alpha$ is relatively open in $S_\alpha$ for each $\alpha\in M$, and that $K\subseteq S$ is compact if and only if $K_\alpha$ is compact for each $\alpha\in M$, and nonempty for at most a finite number of $\alpha\in M$. I can show that $\Sigma$ is a $\sigma$-algebra in $S$ (and, in fact contains the Borel subsets of $S$), and that $\mu$ is a positive measure on $\Sigma$, which is inner regular. I can't seem to prove that $\mu$ is outer regular. Here is the proof in the text: "It is readily seen that $(S,\Sigma,\mu)$ is a regular measure space."
So what's the hangup? Well, let $e\in\Sigma$. If $\mu(e)=\infty$, there is no problem since all open $V$ containing $e$ will have $\mu(V)=\infty$ as well. So suppose $\mu(e)=\sum_{\alpha\in M}\mu_\alpha(e_\alpha)<\infty$. Then $\mu_\alpha(e_\alpha)>0$ for at most a countably infinite number of $\alpha$, say $\alpha\in\{\alpha_1,\alpha_2,\dots\}$. For all the rest of the $\alpha\in M$, we have $\mu_\alpha(e_\alpha)=0$. Now given $\epsilon>0$ we have to find an open set $V$ in $S$ which contains $e$ and which has $\mu(V)=\sum_{\alpha\in M}\mu_\alpha(V_\alpha)<\mu(e)+\epsilon$. My problem is with all those $\alpha\in M-\{\alpha_1,\alpha_2,\dots\}$. For each one, we must have $e_\alpha\subseteq V_\alpha$, and $V_\alpha$ needs to be relatively open in $S_\alpha$. Unless I can guarantee that $\mu_\alpha(V_\alpha)=0$ for all but countably many of those $\alpha$, I'm going to wind up with $\sum_{\alpha\in M}\mu_\alpha(V_\alpha)$ not converging (and therefore being infinite). Right now, I don't see how to accomplish this. The only idea that came to mind was to try to show that there must be an open set $U$ in $S$ containing $e$ such that $\mu(U)<\infty$, but I couldn't even do that. Any ideas on how to proceed would be appreciated.