I'm a little confused about a recent discussion, where a one dimensional wave-function $\Psi(x)$ was considered, associated with a state
$$|\Psi\rangle = \int dx \, \Psi(x) |x\rangle .$$
Now consider a projection operator, somehow narrowing x to be between a and b:
$$P = \int_a^b dx \, |x\rangle \langle x | .$$
The resulting state would be
$$P |\Psi\rangle = \int dx \int_a^b \, dx' |x'\rangle \langle x'|x\rangle \Psi(x)= \int_a^b d x' \Psi(x') |x'\rangle . $$
I would say the result of applying this operator is to project onto a reduced subspace of $|x\rangle$, defined by the interval $[a,b]$.
Now the confusion: is that operator at the same time a measurement? If yes, on what? More precise: what is the observable? I think it cannot define a "measurement" because in that case a reduction of the wave function must occur. Here there is just a reduction of the original Hilbert space to a space containing only vectors $|x\rangle$ with $a \le x \le b$, but not a collapse onto a particular wave function. So in my opinion the projection operator P itself can not be regarded as a measurement.
However, from textbook knowledge, after a measurement the state of the system must be an Eigenstate of a Hermitian operator.
Is it correct to say that P is Hermitean and has Eigenvalues 0 and 1, for vectors $\int dx \Psi(x) |x\rangle $ with $\Psi(x)$ vanishing for any x outside or inside $[a,b]$ respectively? Then the measurement we talk about would be the binary decision whether we find $x$ inside or outside $[a,b]$ and the measured value is the binary information "inside" or "outside" ? In my opinion such measurement does not need to end up in a single $|x\rangle$ as the reduced wave function as it would be in the case of position measurement; I can only say that after measurement the resulting state is with a $\Psi(x)$ either completely vanishing inside or outside $[a,b]$.
Can I say that such process of "incomplete" reduction is a measurement at all? I have some doubts, because this kind of measurement would end up in sub state, that is still a superposition of base vectors and does, in general, not reduce to a particular one.
Please correct me where I'm not correct.