Integrating unbounded functions w.r.t. a projection-valued measure I was looking at Frederic Schuller's lectures on quantum theory and there, when defining integrals of unbounded functions with respect to a projection-valued measure, it is left as an exercise to verify the following:
Let $f:\mathbb{R}\to\mathbb{C}$ be an unbounded measurable function, and define
$$f_n:=\chi_{\{|f|\leq n\}}f$$
for $n\in\mathbb{N}_1$. Then $(f_n)_{n=1}^\infty$ is a Cauchy sequence in $L^2(\mathbb{R})$, which then implies that $\left(\int_\mathbb{R}f_n dP\psi\right)_{n=1}^\infty\subset H$ is Cauchy in $H$ for every $\psi\in H$, where $H$ is a Hilbert space.
I don't see why the first claim should hold: For example consider $f=id_\mathbb{R}$. Then, if $n<m$,
$$|f_m-f_n|\ge ||f_m|-|f_n||=|\chi_{\{|f|\leq m\}}|f|-\chi_{\{|f|\leq n\}}|f||=\chi_{\{|f|\in (n,m]\}}|f|,$$
which when keeping any $n$ fixed and letting $m\to\infty$ does not have an $L^2$-norm that is small. If this is false, then how does one show that these integrals make sense for unbounded $f$?
 A: This statement is wrong. In fact, since $f_n\to f$ a.e. and convergence in $L^2$ implies a.e. convergence along a subsequence, the sequence $(f_n)$ can only be Cauchy in $L^2$ if $f$ is in $L^2$. Of course, in general, $f_n$ is not even an element of $L^2$, so that $(f_n)$ has no chance of being a Cauchy sequence in $L^2$. Likewise, the integral with respect to a projection-valued measure is not defined for all unbounded functions.
What is true is the following: If $f\in L^2(\mathbb R,\mu)$ - for the applications to the spectral theorem you really need to consider arbitrary (finite Borel) measures -, then $(f_n)$ is Cauchy in $L^2(\mathbb R,\mu)$. In fact, the convergence of $(f_n)$ to $f$ in $L^2(\mathbb R,\mu)$ follows from an application of the dominated convergence theorem.
This can be used to show the existence of $\int f\,d P\psi$ whenever $f\in L^2(\mathbb R,\mu_\psi)$, where $\mu_\psi$ is the (scalar) measure given by $\mu_\psi(A)=\lVert P(A)\psi\rVert^2$. The set of such $\psi$ then forms the domain of $\int f\,dP\psi$.
