Spectral Measures: Pushforward

Consider the pushforward: $$E_\eta:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E_\eta(A):=E(\eta^{-1}A)$$

Their domain agree as:* $$\int|\vartheta\circ\eta|^2\mathrm{d}\nu_\varphi(\lambda)=\int|\vartheta|^2\mathrm{d}\nu^\eta_\varphi$$

Denote for shorthand: $$\mathcal{D}:=\mathcal{D}(\vartheta\circ\eta)(N)=\mathcal{D}(\vartheta\circ\eta)(E)=\mathcal{D}\vartheta(E_\eta)$$

In particular one has:** $$\varphi\in\mathcal{D}:\quad\int|\vartheta\circ\eta|\,\mathrm{d}|\mu_{\varphi\chi}|<\infty\quad\int|\vartheta|\,\mathrm{d}|\mu^\eta_{\varphi\chi}|<\infty$$

For simple functions: $$\int s\circ\eta\,\mathrm{d}\mu_{\varphi\chi}=\sum_kb_k\mu(\eta^{-1}B_k)=\int s\,\mathrm{d}\mu^\eta_{\varphi\chi}$$

By Lebesgue one gets:*** $$\varphi\in\mathcal{D}:\quad\int\vartheta\circ\eta\,\mathrm{d}\mu_{\varphi\chi}=\int\vartheta\,\mathrm{d}\mu^\eta_{\varphi\chi}(\lambda)$$

But note that it was: $$\mathrm{id}(E_\eta)=\eta(E)=\eta(N)$$

Concluding the assertion.

*See the thread: Pushforward (BM)

**Note the thread: Pushforward (CM)

***See the thread: Lebesgue (CM)