Example. Integrable observables need not form an algebra [fdip-0030]

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Let \(Q\) be positive self-adjoint on \((E,h)\) with \(n=\dim E>0\). The Borel function \(F=e^{S_Q/2}\) is integrable under the Gaussian law, since \[\int _E Fw_Q\,d\lambda _E =\int _E e^{-S_Q/2}\,d\lambda _E=Z_{Q/2}<\infty \] by positive normalization. But \(F^2=e^{S_Q}\) is not integrable: \(F^2w_Q=1\), and \(\lambda _E(E)=\infty \). Indeed orthonormal cubes have masses \((2R)^n\) tending to infinity.

Thus actual integrable observables form a vector space but are not closed under multiplication. The same counterexample works for their almost-everywhere classes in \(L^1(\mu _Q)\). The polynomial algebra is an integrable algebra by Proposition [fdip-002W]. For \(n=0\), every real observable is constant on a singleton, so this failure is absent.

This is a direct comparison of the two normalizers, using Theorem [fdip-002T] and the scalar ingredient in Keith Conrad's [conradGaussianIntegral, Section 2, PDF pp. 1-2]; it also retains \(e^{S_Q}\) as a concrete Borel nonintegrable observable in positive dimension.