Let $\mathscr{H}$ be a Hilbert space, $\psi \in \mathscr{H}$ and $A$ be a densely-defined self-adjoint operator. From the Borel functional calculus, for each measurable function $f: \mathbb{R} \to \mathbb{C}$, there exists a bounded linear operator $f(A)$ on $\mathscr{H}$. For each bounded continuous function $f: \mathbb{R}\to \mathbb{C}$, define: $$\omega(f) = \langle \psi, f(A)\psi\rangle$$ This is a positive linear functional, so by Riesz-Markov Theorem there exists a Borel measure $\mu_{\psi}$ on $\mathbb{R}$ such that: $$\langle \psi, f(A)\psi \rangle = \int_{\mathbb{R}}f(x)d\mu_{\psi}(x).$$
Question: Suppose $\mu_{\psi} = \mu_{\varphi}$. Is it true that $\psi = \varphi$? I could not conclude this from $\langle \psi,f(A)\rangle = \langle \varphi,f(A)\varphi\rangle$.
The claim is not true even for two-dimensional space with orthonormal basis $e_1,e_2$ and the identity operator. Then the measures corresponding to $e_1$ and $e_2$ are equal, although the vectors are linearly independent. I guess the conclusion may hold for operators with simple spectrum