Spectral Measures: Uniqueness Given a Hilbert space $\mathcal{H}$.
Consider spectral measures:
$$E^{(\prime)}:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$
Denote their operators by:
$$M^{(\prime)}:=\int\lambda\mathrm{d}E^{(\prime)}(\lambda)$$
Then one has:
$$M=M'\iff E=E'$$
How can I prove this?
 A: I finally found a proof. :D
Consider the projections:
$$N_\Re:=\frac{1}{2}\{N+N^*\}=\frac{1}{2}\{N'+N'^*\}=:N'_\Re$$
$$N_\Im:=\frac{1}{2i}\{N-N^*\}=\frac{1}{2i}\{N'-N'^*\}=:N'_\Im$$
Their resolvents agree:
$$R_\alpha(z)=(z-N_\alpha)^{-1}=(z-N'_\alpha)^{-1}=R'_\alpha(z)$$
By Stone's formula:
$$E_\alpha(-\infty,\lambda_\alpha]\varphi=\lim_{\delta\to0^+}\lim_{\varepsilon\to0^+}\int_{-\infty}^{\lambda_\alpha+\delta}\Delta R_\alpha(s\pm i\varepsilon)\varphi\mathrm{d}s\\
=\lim_{\delta\to0^+}\lim_{\varepsilon\to0^+}\int_{-\infty}^{\lambda_\alpha+\delta}\Delta R'_\alpha(s\pm i\varepsilon)\varphi\mathrm{d}s=E'_\alpha(-\infty,\lambda_\alpha]\varphi$$
By construction one has:
$$E^{(\prime)}_\Re(A)=E^{(\prime)}(A\times\mathbb{R})\quad E^{(\prime)}_\Im(B)=E^{(\prime)}(\mathbb{R}\times B)$$
Therefore one obtains:
$$E(-\infty,\lambda]=E\bigg((-\infty,\lambda_\Re]\times\mathbb{R}\bigg)E\bigg(\mathbb{R}\times(-\infty,\lambda_\Im]\bigg)\\
=E'\bigg((-\infty,\lambda_\Re]\times\mathbb{R}\bigg)E'\bigg(\mathbb{R}\times(-\infty,\lambda_\Im]\bigg)=E'(-\infty,\lambda]$$
Concluding the assertion.
