Let $H$ be a complex Hilbert space (not necessary separable).

Spectral Theorem: Let $A_1$ and $A_2$ be two commuting normal operators, then there exists a measure space $(X,\mathcal{E},\mu)$, two functions $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:H\longrightarrow L^2(\mu)$, such that each $A_k$ is unitarily equivalent to multiplication by $\varphi_k$, $k=1,2$. i.e. $$UA_kU^*f=\varphi_kf,\;\forall f\in H,\,k=1,2.$$

Is $\mu$ semifinite? i.e. for each $E \in \mathcal{E}$ with $\mu(E) = \infty$ , there exists $F \subset E$ and $F \in \mathcal{E}$ and $0 < \mu(F) < \infty$.

If $H$ is a separable complex Hilbert space, then $(X,\mathcal{E},\mu)$ is a $\sigma$-finite measure space and so $\mu$ is semifinite.

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    $\begingroup$ If $\mu$ has an infinite atom, then every $L^2$ function must vanish on that set, so it might as well not be there at all. This suggests that you should be able to replace $\mu$ by its semifinite part and get the same $L^2$ space, though I have not checked the details. $\endgroup$ – Nate Eldredge Mar 7 '18 at 15:30
  • $\begingroup$ Sorry, wrong link: I meant math.stackexchange.com/questions/393985/… $\endgroup$ – Nate Eldredge Mar 7 '18 at 15:39
  • $\begingroup$ @NateEldredge When we remplace $\mu$ by its semifinite part that doen't mean that $\mu$ is semifinite. Do you agree with me? $\endgroup$ – Student Mar 9 '18 at 5:06
  • $\begingroup$ What I am asserting is that there always exists a semifinite measure satisfying the given properties. Of course, there will always exist non-semifinite ones as well (take any such measure and if it's semifinite then consider a space with one additional point that has measure infinity). My suspicion is that you can prove there's a semifinite one by saying "let $\mu$ be a measure satisfying the conclusion of the spectral theorem; if it is not semifinite, let $\mu_0$ be its semifinite part and then $\mu_0$ satisfies." But I can't quite finish the proof of this. $\endgroup$ – Nate Eldredge Mar 9 '18 at 16:06
  • $\begingroup$ @NateEldredge Yes I understand and thank. The answer of Nik Weaver says that $\mu$ is localizable but I don't understand very well the difference between localizable and semifinite. $\endgroup$ – Student Mar 9 '18 at 16:10

The answer is yes. Let me explain why.

Let $v \in E$ be any nonzero vector. Then define $E_0$ to be the closure of the set of vectors of the form $p(A_1,A_1^*, A_2, A_2^*)v$ where $p$ is a complex polynomial. By restricting $p$ to have rational coefficients you can see that $E_0$ is separable. Now the point is that $A_1E_0 \subseteq E_0$ and $A_1E_0^\perp \subseteq E_0^\perp$, and similarly for $A_1^*$, $A_2$, and $A_2^*$. I.e., $E_0$ is a "reducing" subspace. Now you can Zornicate and get a family of mutually orthogonal separable subspaces $E_\alpha$, each of which is reducing for $A_1$ and $A_2$, and which span all of $E$. Apply the spectral theorem separately on each $E_\alpha$.

This shows you something even stronger than semifiniteness --- it shows you that $\mu$ is localizable: the measure space can be decomposed into a disjoint family of finite measure subsets, such that the measure of any set is the sum of the measures of its intersections with the members of this family.

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  • $\begingroup$ Thank you for your answer but I don't understand why $A_1E_0 \subseteq E_0$ and $A_1E_0^\perp \subseteq E_0^\perp$? $\endgroup$ – Student Mar 8 '18 at 15:59
  • $\begingroup$ Well, these are exercises. $E_0$ is the smallest subspace which is closed in norm and closed under application of $A_1$, $A_1^*$, $A_2$, and $A_2^*$. $\endgroup$ – Nik Weaver Mar 8 '18 at 16:16
  • $\begingroup$ Thank you. I think that a localizable measure is not in general $\sigma$-finite. Do you agree with me? $\endgroup$ – Student Mar 8 '18 at 20:16
  • $\begingroup$ No, for example counting measure on any set is localizable. It is a generalization of $\sigma$-finiteness to the nonseparable setting. Here is a little primer on localizablility for you (from a book of mine which is in press). $\endgroup$ – Nik Weaver Mar 8 '18 at 20:50
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    $\begingroup$ To be pedantic, I guess we should say that $\mu$ can be taken localizable. There is not a unique $\mu$ satisfying the theorem, and there will certainly exist such $\mu$ that are not semifinite. $\endgroup$ – Nate Eldredge Mar 9 '18 at 16:08

If we replace $\mu$ by its semifinite part $\mu_1$ defined as $$ \mu_{1}(E) = \sup\{\mu(A) \mid A \subseteq E \text{ measurable, } \mu(A)< \infty\} \quad \text{for }E \in \mathcal{E}, $$ we have by the following answer (1): $$L^2(\mu)"=" L^2(\mu_1).$$

Thus $\mu$ can be taken semifinite measure.

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